Non zero solution of $3xcos(x) + (-3 + x^2)sin(x)=0$












1












$begingroup$


How can I find exact non-zero solution of $3xcos(x) + (-3 + x^2)sin(x)=0$. Simple analysis and the below plot show that the equation has an infinite number of non-zero solutions.



enter image description here






calculus trigonometry trigonometric-series elementary-functions







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    How can I find exact non-zero solution of $3xcos(x) + (-3 + x^2)sin(x)=0$. Simple analysis and the below plot show that the equation has an infinite number of non-zero solutions.



    enter image description here






    calculus trigonometry trigonometric-series elementary-functions







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      How can I find exact non-zero solution of $3xcos(x) + (-3 + x^2)sin(x)=0$. Simple analysis and the below plot show that the equation has an infinite number of non-zero solutions.



      enter image description here






      calculus trigonometry trigonometric-series elementary-functions







      share|cite|improve this question











      $endgroup$




      How can I find exact non-zero solution of $3xcos(x) + (-3 + x^2)sin(x)=0$. Simple analysis and the below plot show that the equation has an infinite number of non-zero solutions.



      enter image description here






      calculus trigonometry trigonometric-series elementary-functions




      calculus trigonometry trigonometric-series elementary-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 23:23









      David G. Stork

      11k41432




      11k41432










      asked Jan 7 at 17:39









      ershersh

      408113




      408113






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          There is an infinite number of solutions. Choose one region and find a solution there:



          FindMinimum[Abs[3 x Cos[x] + (x^2 - 3) Sin[x]], {x, 9.5}] // Quiet


          ${text{1.6217274456664654$grave{ }$*${}^{wedge}$-7},{xto 5.76346}}$



          enter image description here






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I need an exact closed form solution, which could be in the form of a series.
            $endgroup$
            – ersh
            Jan 7 at 18:01








          • 1




            $begingroup$
            I formed an algebraic expression for tan(x), and then used it in conjuction with the taylor series for arctan(x), but couldn't get anywhere. I think the solution lies in exploring the context of the question more deeply. If this is a problem from a textbook, you would attempt to use the methods introduced in the pages leading up to the problem. Failing that, another approach would be to try to find $f(x)$ so that some clever manipulation of either $d[f(x)]/dx$ or $int f(x)dx$ helps you to achieve a closed form solution.
            $endgroup$
            – user2661923
            Jan 7 at 20:19






          • 1




            $begingroup$
            @DavidG.Stork a 2nd try is $3x cos(x) = (3-x^2)sin(x).$ Square both sides and substitute $1 - cos^2(x) = sin^2(x).$ This leads to an equation like $(x^4 + ax^2 + b)(cos^2(x) = (x^4 + cx^2 + d).$ Unfortunately, I can't see anyway to use such an equation, and I don't see how (for example) attempting to combine the taylor series for cosine with the equation would help.
            $endgroup$
            – user2661923
            Jan 7 at 20:31






          • 1




            $begingroup$
            @user2661923: No amount of manipulation can alter the topological/algebraic fact that there is an infinite number of solutions to this problem.
            $endgroup$
            – David G. Stork
            Jan 7 at 23:22








          • 1




            $begingroup$
            @DavidG.Stork agreed, but that fact does not imply that the problem won't succumb to taylor series expansion or manipulation of $d[f(x)]/dx$ or $int f(x)dx.$ For example, the taylor series for $cos(x)$ converges to 0 at an infinite number of points.
            $endgroup$
            – user2661923
            Jan 7 at 23:59



















          1












          $begingroup$

          I am certain that
          there is no exact solution to this.
          However,
          by looking at regions where either
          $sin(x)$ or $cos(x)$ is small,
          I think that it can be shown that
          $f(x)
          =3xcos(x) + (-3 + x^2)sin(x)
          $
          will change sign in these regions.



          In particular,
          let $x = npi+y$
          where $y$ is small.



          Then
          $sin(x) = sin(y)
          approx y$          .



          Also,
          $cos(x)
          =sqrt{1-sin^2(x)}
          =sqrt{1-sin^2(y)}
          approxsqrt{1-y^2}
          approx 1-frac{y^2}{2}
          $          .



          Similarly,
          if $x = (n+frac12)pi+y$
          where $y$ is small.



          Then
          $cos(x) = sin(frac{pi}{2}-x)
          approx y$
          and
          $sin(x)
          approx 1-frac{y^2}{2}
          $          .



          Try each of these in $f(x)$
          and see which can be made small
          by making $y$ small.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            There's probably no closed-form solution for the real roots you're looking for, although you have a chance with the complex solutions.



            Let's notice that $3x cos x$ and $left( x^2 - 3 right) sin x$ have the same trigonometric argument. That means that the Harmonic Addition Theorem can be used: we can rewrite this as one trigonometric function.



            HAT states that
            begin{align}a cos x + b sin x &= A cos left(x + B right),
            end{align}
            where $a^2 + b^2 = A^2$ and $-frac{b}a = tan B$.



            So, substituting $3x cos x$ for $a$ and $x^2 - 3$ for $b$, we find that $A^2 = x^4 + 3x^2 + 9$ and $frac{3 - x^2}{3x} = tan B$.



            This of course has infinitely many solutions, although the factor of $A$ only contributes some complex roots. Ignoring questions of signs and branch choice, let's declare that $A = sqrt{x^4 + 3x^2 + 9}$ and $B = tan^{-1} frac{3 - x^2}{3x}$, or in other words
            $$3x cos x + (x^2 - 3) sin x = sqrt{x^4 + 3x^2 + 9} cos left( x + tan^{-1} frac{3 - x^2}{3x} right).$$



            The problem can therefore be rewritten as finding the zeroes of $sqrt{x^4 + 3x^2 + 9}$ and $cos left( x + tan^{-1} frac{3 - x^2}{3x} right)$, although I doubt you'll find any good closed form for the latter.



            This WolframAlpha plot looks like it agrees with your image.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for such a nice manipulation. But still the expression $sqrt(x^4+3x^2+9)=0$ doesn't have the reql solutions. I will try to work with other factor
              $endgroup$
              – ersh
              Jan 8 at 5:05











            Your Answer





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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            There is an infinite number of solutions. Choose one region and find a solution there:



            FindMinimum[Abs[3 x Cos[x] + (x^2 - 3) Sin[x]], {x, 9.5}] // Quiet


            ${text{1.6217274456664654$grave{ }$*${}^{wedge}$-7},{xto 5.76346}}$



            enter image description here






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I need an exact closed form solution, which could be in the form of a series.
              $endgroup$
              – ersh
              Jan 7 at 18:01








            • 1




              $begingroup$
              I formed an algebraic expression for tan(x), and then used it in conjuction with the taylor series for arctan(x), but couldn't get anywhere. I think the solution lies in exploring the context of the question more deeply. If this is a problem from a textbook, you would attempt to use the methods introduced in the pages leading up to the problem. Failing that, another approach would be to try to find $f(x)$ so that some clever manipulation of either $d[f(x)]/dx$ or $int f(x)dx$ helps you to achieve a closed form solution.
              $endgroup$
              – user2661923
              Jan 7 at 20:19






            • 1




              $begingroup$
              @DavidG.Stork a 2nd try is $3x cos(x) = (3-x^2)sin(x).$ Square both sides and substitute $1 - cos^2(x) = sin^2(x).$ This leads to an equation like $(x^4 + ax^2 + b)(cos^2(x) = (x^4 + cx^2 + d).$ Unfortunately, I can't see anyway to use such an equation, and I don't see how (for example) attempting to combine the taylor series for cosine with the equation would help.
              $endgroup$
              – user2661923
              Jan 7 at 20:31






            • 1




              $begingroup$
              @user2661923: No amount of manipulation can alter the topological/algebraic fact that there is an infinite number of solutions to this problem.
              $endgroup$
              – David G. Stork
              Jan 7 at 23:22








            • 1




              $begingroup$
              @DavidG.Stork agreed, but that fact does not imply that the problem won't succumb to taylor series expansion or manipulation of $d[f(x)]/dx$ or $int f(x)dx.$ For example, the taylor series for $cos(x)$ converges to 0 at an infinite number of points.
              $endgroup$
              – user2661923
              Jan 7 at 23:59
















            1












            $begingroup$

            There is an infinite number of solutions. Choose one region and find a solution there:



            FindMinimum[Abs[3 x Cos[x] + (x^2 - 3) Sin[x]], {x, 9.5}] // Quiet


            ${text{1.6217274456664654$grave{ }$*${}^{wedge}$-7},{xto 5.76346}}$



            enter image description here






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I need an exact closed form solution, which could be in the form of a series.
              $endgroup$
              – ersh
              Jan 7 at 18:01








            • 1




              $begingroup$
              I formed an algebraic expression for tan(x), and then used it in conjuction with the taylor series for arctan(x), but couldn't get anywhere. I think the solution lies in exploring the context of the question more deeply. If this is a problem from a textbook, you would attempt to use the methods introduced in the pages leading up to the problem. Failing that, another approach would be to try to find $f(x)$ so that some clever manipulation of either $d[f(x)]/dx$ or $int f(x)dx$ helps you to achieve a closed form solution.
              $endgroup$
              – user2661923
              Jan 7 at 20:19






            • 1




              $begingroup$
              @DavidG.Stork a 2nd try is $3x cos(x) = (3-x^2)sin(x).$ Square both sides and substitute $1 - cos^2(x) = sin^2(x).$ This leads to an equation like $(x^4 + ax^2 + b)(cos^2(x) = (x^4 + cx^2 + d).$ Unfortunately, I can't see anyway to use such an equation, and I don't see how (for example) attempting to combine the taylor series for cosine with the equation would help.
              $endgroup$
              – user2661923
              Jan 7 at 20:31






            • 1




              $begingroup$
              @user2661923: No amount of manipulation can alter the topological/algebraic fact that there is an infinite number of solutions to this problem.
              $endgroup$
              – David G. Stork
              Jan 7 at 23:22








            • 1




              $begingroup$
              @DavidG.Stork agreed, but that fact does not imply that the problem won't succumb to taylor series expansion or manipulation of $d[f(x)]/dx$ or $int f(x)dx.$ For example, the taylor series for $cos(x)$ converges to 0 at an infinite number of points.
              $endgroup$
              – user2661923
              Jan 7 at 23:59














            1












            1








            1





            $begingroup$

            There is an infinite number of solutions. Choose one region and find a solution there:



            FindMinimum[Abs[3 x Cos[x] + (x^2 - 3) Sin[x]], {x, 9.5}] // Quiet


            ${text{1.6217274456664654$grave{ }$*${}^{wedge}$-7},{xto 5.76346}}$



            enter image description here






            share|cite|improve this answer









            $endgroup$



            There is an infinite number of solutions. Choose one region and find a solution there:



            FindMinimum[Abs[3 x Cos[x] + (x^2 - 3) Sin[x]], {x, 9.5}] // Quiet


            ${text{1.6217274456664654$grave{ }$*${}^{wedge}$-7},{xto 5.76346}}$



            enter image description here







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 7 at 17:46









            David G. StorkDavid G. Stork

            11k41432




            11k41432












            • $begingroup$
              I need an exact closed form solution, which could be in the form of a series.
              $endgroup$
              – ersh
              Jan 7 at 18:01








            • 1




              $begingroup$
              I formed an algebraic expression for tan(x), and then used it in conjuction with the taylor series for arctan(x), but couldn't get anywhere. I think the solution lies in exploring the context of the question more deeply. If this is a problem from a textbook, you would attempt to use the methods introduced in the pages leading up to the problem. Failing that, another approach would be to try to find $f(x)$ so that some clever manipulation of either $d[f(x)]/dx$ or $int f(x)dx$ helps you to achieve a closed form solution.
              $endgroup$
              – user2661923
              Jan 7 at 20:19






            • 1




              $begingroup$
              @DavidG.Stork a 2nd try is $3x cos(x) = (3-x^2)sin(x).$ Square both sides and substitute $1 - cos^2(x) = sin^2(x).$ This leads to an equation like $(x^4 + ax^2 + b)(cos^2(x) = (x^4 + cx^2 + d).$ Unfortunately, I can't see anyway to use such an equation, and I don't see how (for example) attempting to combine the taylor series for cosine with the equation would help.
              $endgroup$
              – user2661923
              Jan 7 at 20:31






            • 1




              $begingroup$
              @user2661923: No amount of manipulation can alter the topological/algebraic fact that there is an infinite number of solutions to this problem.
              $endgroup$
              – David G. Stork
              Jan 7 at 23:22








            • 1




              $begingroup$
              @DavidG.Stork agreed, but that fact does not imply that the problem won't succumb to taylor series expansion or manipulation of $d[f(x)]/dx$ or $int f(x)dx.$ For example, the taylor series for $cos(x)$ converges to 0 at an infinite number of points.
              $endgroup$
              – user2661923
              Jan 7 at 23:59


















            • $begingroup$
              I need an exact closed form solution, which could be in the form of a series.
              $endgroup$
              – ersh
              Jan 7 at 18:01








            • 1




              $begingroup$
              I formed an algebraic expression for tan(x), and then used it in conjuction with the taylor series for arctan(x), but couldn't get anywhere. I think the solution lies in exploring the context of the question more deeply. If this is a problem from a textbook, you would attempt to use the methods introduced in the pages leading up to the problem. Failing that, another approach would be to try to find $f(x)$ so that some clever manipulation of either $d[f(x)]/dx$ or $int f(x)dx$ helps you to achieve a closed form solution.
              $endgroup$
              – user2661923
              Jan 7 at 20:19






            • 1




              $begingroup$
              @DavidG.Stork a 2nd try is $3x cos(x) = (3-x^2)sin(x).$ Square both sides and substitute $1 - cos^2(x) = sin^2(x).$ This leads to an equation like $(x^4 + ax^2 + b)(cos^2(x) = (x^4 + cx^2 + d).$ Unfortunately, I can't see anyway to use such an equation, and I don't see how (for example) attempting to combine the taylor series for cosine with the equation would help.
              $endgroup$
              – user2661923
              Jan 7 at 20:31






            • 1




              $begingroup$
              @user2661923: No amount of manipulation can alter the topological/algebraic fact that there is an infinite number of solutions to this problem.
              $endgroup$
              – David G. Stork
              Jan 7 at 23:22








            • 1




              $begingroup$
              @DavidG.Stork agreed, but that fact does not imply that the problem won't succumb to taylor series expansion or manipulation of $d[f(x)]/dx$ or $int f(x)dx.$ For example, the taylor series for $cos(x)$ converges to 0 at an infinite number of points.
              $endgroup$
              – user2661923
              Jan 7 at 23:59
















            $begingroup$
            I need an exact closed form solution, which could be in the form of a series.
            $endgroup$
            – ersh
            Jan 7 at 18:01






            $begingroup$
            I need an exact closed form solution, which could be in the form of a series.
            $endgroup$
            – ersh
            Jan 7 at 18:01






            1




            1




            $begingroup$
            I formed an algebraic expression for tan(x), and then used it in conjuction with the taylor series for arctan(x), but couldn't get anywhere. I think the solution lies in exploring the context of the question more deeply. If this is a problem from a textbook, you would attempt to use the methods introduced in the pages leading up to the problem. Failing that, another approach would be to try to find $f(x)$ so that some clever manipulation of either $d[f(x)]/dx$ or $int f(x)dx$ helps you to achieve a closed form solution.
            $endgroup$
            – user2661923
            Jan 7 at 20:19




            $begingroup$
            I formed an algebraic expression for tan(x), and then used it in conjuction with the taylor series for arctan(x), but couldn't get anywhere. I think the solution lies in exploring the context of the question more deeply. If this is a problem from a textbook, you would attempt to use the methods introduced in the pages leading up to the problem. Failing that, another approach would be to try to find $f(x)$ so that some clever manipulation of either $d[f(x)]/dx$ or $int f(x)dx$ helps you to achieve a closed form solution.
            $endgroup$
            – user2661923
            Jan 7 at 20:19




            1




            1




            $begingroup$
            @DavidG.Stork a 2nd try is $3x cos(x) = (3-x^2)sin(x).$ Square both sides and substitute $1 - cos^2(x) = sin^2(x).$ This leads to an equation like $(x^4 + ax^2 + b)(cos^2(x) = (x^4 + cx^2 + d).$ Unfortunately, I can't see anyway to use such an equation, and I don't see how (for example) attempting to combine the taylor series for cosine with the equation would help.
            $endgroup$
            – user2661923
            Jan 7 at 20:31




            $begingroup$
            @DavidG.Stork a 2nd try is $3x cos(x) = (3-x^2)sin(x).$ Square both sides and substitute $1 - cos^2(x) = sin^2(x).$ This leads to an equation like $(x^4 + ax^2 + b)(cos^2(x) = (x^4 + cx^2 + d).$ Unfortunately, I can't see anyway to use such an equation, and I don't see how (for example) attempting to combine the taylor series for cosine with the equation would help.
            $endgroup$
            – user2661923
            Jan 7 at 20:31




            1




            1




            $begingroup$
            @user2661923: No amount of manipulation can alter the topological/algebraic fact that there is an infinite number of solutions to this problem.
            $endgroup$
            – David G. Stork
            Jan 7 at 23:22






            $begingroup$
            @user2661923: No amount of manipulation can alter the topological/algebraic fact that there is an infinite number of solutions to this problem.
            $endgroup$
            – David G. Stork
            Jan 7 at 23:22






            1




            1




            $begingroup$
            @DavidG.Stork agreed, but that fact does not imply that the problem won't succumb to taylor series expansion or manipulation of $d[f(x)]/dx$ or $int f(x)dx.$ For example, the taylor series for $cos(x)$ converges to 0 at an infinite number of points.
            $endgroup$
            – user2661923
            Jan 7 at 23:59




            $begingroup$
            @DavidG.Stork agreed, but that fact does not imply that the problem won't succumb to taylor series expansion or manipulation of $d[f(x)]/dx$ or $int f(x)dx.$ For example, the taylor series for $cos(x)$ converges to 0 at an infinite number of points.
            $endgroup$
            – user2661923
            Jan 7 at 23:59











            1












            $begingroup$

            I am certain that
            there is no exact solution to this.
            However,
            by looking at regions where either
            $sin(x)$ or $cos(x)$ is small,
            I think that it can be shown that
            $f(x)
            =3xcos(x) + (-3 + x^2)sin(x)
            $
            will change sign in these regions.



            In particular,
            let $x = npi+y$
            where $y$ is small.



            Then
            $sin(x) = sin(y)
            approx y$            .



            Also,
            $cos(x)
            =sqrt{1-sin^2(x)}
            =sqrt{1-sin^2(y)}
            approxsqrt{1-y^2}
            approx 1-frac{y^2}{2}
            $            .



            Similarly,
            if $x = (n+frac12)pi+y$
            where $y$ is small.



            Then
            $cos(x) = sin(frac{pi}{2}-x)
            approx y$
            and
            $sin(x)
            approx 1-frac{y^2}{2}
            $            .



            Try each of these in $f(x)$
            and see which can be made small
            by making $y$ small.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              I am certain that
              there is no exact solution to this.
              However,
              by looking at regions where either
              $sin(x)$ or $cos(x)$ is small,
              I think that it can be shown that
              $f(x)
              =3xcos(x) + (-3 + x^2)sin(x)
              $
              will change sign in these regions.



              In particular,
              let $x = npi+y$
              where $y$ is small.



              Then
              $sin(x) = sin(y)
              approx y$              .



              Also,
              $cos(x)
              =sqrt{1-sin^2(x)}
              =sqrt{1-sin^2(y)}
              approxsqrt{1-y^2}
              approx 1-frac{y^2}{2}
              $              .



              Similarly,
              if $x = (n+frac12)pi+y$
              where $y$ is small.



              Then
              $cos(x) = sin(frac{pi}{2}-x)
              approx y$
              and
              $sin(x)
              approx 1-frac{y^2}{2}
              $              .



              Try each of these in $f(x)$
              and see which can be made small
              by making $y$ small.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                I am certain that
                there is no exact solution to this.
                However,
                by looking at regions where either
                $sin(x)$ or $cos(x)$ is small,
                I think that it can be shown that
                $f(x)
                =3xcos(x) + (-3 + x^2)sin(x)
                $
                will change sign in these regions.



                In particular,
                let $x = npi+y$
                where $y$ is small.



                Then
                $sin(x) = sin(y)
                approx y$                .



                Also,
                $cos(x)
                =sqrt{1-sin^2(x)}
                =sqrt{1-sin^2(y)}
                approxsqrt{1-y^2}
                approx 1-frac{y^2}{2}
                $                .



                Similarly,
                if $x = (n+frac12)pi+y$
                where $y$ is small.



                Then
                $cos(x) = sin(frac{pi}{2}-x)
                approx y$
                and
                $sin(x)
                approx 1-frac{y^2}{2}
                $                .



                Try each of these in $f(x)$
                and see which can be made small
                by making $y$ small.






                share|cite|improve this answer









                $endgroup$



                I am certain that
                there is no exact solution to this.
                However,
                by looking at regions where either
                $sin(x)$ or $cos(x)$ is small,
                I think that it can be shown that
                $f(x)
                =3xcos(x) + (-3 + x^2)sin(x)
                $
                will change sign in these regions.



                In particular,
                let $x = npi+y$
                where $y$ is small.



                Then
                $sin(x) = sin(y)
                approx y$                .



                Also,
                $cos(x)
                =sqrt{1-sin^2(x)}
                =sqrt{1-sin^2(y)}
                approxsqrt{1-y^2}
                approx 1-frac{y^2}{2}
                $                .



                Similarly,
                if $x = (n+frac12)pi+y$
                where $y$ is small.



                Then
                $cos(x) = sin(frac{pi}{2}-x)
                approx y$
                and
                $sin(x)
                approx 1-frac{y^2}{2}
                $                .



                Try each of these in $f(x)$
                and see which can be made small
                by making $y$ small.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 1:19









                marty cohenmarty cohen

                73.7k549128




                73.7k549128























                    1












                    $begingroup$

                    There's probably no closed-form solution for the real roots you're looking for, although you have a chance with the complex solutions.



                    Let's notice that $3x cos x$ and $left( x^2 - 3 right) sin x$ have the same trigonometric argument. That means that the Harmonic Addition Theorem can be used: we can rewrite this as one trigonometric function.



                    HAT states that
                    begin{align}a cos x + b sin x &= A cos left(x + B right),
                    end{align}
                    where $a^2 + b^2 = A^2$ and $-frac{b}a = tan B$.



                    So, substituting $3x cos x$ for $a$ and $x^2 - 3$ for $b$, we find that $A^2 = x^4 + 3x^2 + 9$ and $frac{3 - x^2}{3x} = tan B$.



                    This of course has infinitely many solutions, although the factor of $A$ only contributes some complex roots. Ignoring questions of signs and branch choice, let's declare that $A = sqrt{x^4 + 3x^2 + 9}$ and $B = tan^{-1} frac{3 - x^2}{3x}$, or in other words
                    $$3x cos x + (x^2 - 3) sin x = sqrt{x^4 + 3x^2 + 9} cos left( x + tan^{-1} frac{3 - x^2}{3x} right).$$



                    The problem can therefore be rewritten as finding the zeroes of $sqrt{x^4 + 3x^2 + 9}$ and $cos left( x + tan^{-1} frac{3 - x^2}{3x} right)$, although I doubt you'll find any good closed form for the latter.



                    This WolframAlpha plot looks like it agrees with your image.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thanks for such a nice manipulation. But still the expression $sqrt(x^4+3x^2+9)=0$ doesn't have the reql solutions. I will try to work with other factor
                      $endgroup$
                      – ersh
                      Jan 8 at 5:05
















                    1












                    $begingroup$

                    There's probably no closed-form solution for the real roots you're looking for, although you have a chance with the complex solutions.



                    Let's notice that $3x cos x$ and $left( x^2 - 3 right) sin x$ have the same trigonometric argument. That means that the Harmonic Addition Theorem can be used: we can rewrite this as one trigonometric function.



                    HAT states that
                    begin{align}a cos x + b sin x &= A cos left(x + B right),
                    end{align}
                    where $a^2 + b^2 = A^2$ and $-frac{b}a = tan B$.



                    So, substituting $3x cos x$ for $a$ and $x^2 - 3$ for $b$, we find that $A^2 = x^4 + 3x^2 + 9$ and $frac{3 - x^2}{3x} = tan B$.



                    This of course has infinitely many solutions, although the factor of $A$ only contributes some complex roots. Ignoring questions of signs and branch choice, let's declare that $A = sqrt{x^4 + 3x^2 + 9}$ and $B = tan^{-1} frac{3 - x^2}{3x}$, or in other words
                    $$3x cos x + (x^2 - 3) sin x = sqrt{x^4 + 3x^2 + 9} cos left( x + tan^{-1} frac{3 - x^2}{3x} right).$$



                    The problem can therefore be rewritten as finding the zeroes of $sqrt{x^4 + 3x^2 + 9}$ and $cos left( x + tan^{-1} frac{3 - x^2}{3x} right)$, although I doubt you'll find any good closed form for the latter.



                    This WolframAlpha plot looks like it agrees with your image.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thanks for such a nice manipulation. But still the expression $sqrt(x^4+3x^2+9)=0$ doesn't have the reql solutions. I will try to work with other factor
                      $endgroup$
                      – ersh
                      Jan 8 at 5:05














                    1












                    1








                    1





                    $begingroup$

                    There's probably no closed-form solution for the real roots you're looking for, although you have a chance with the complex solutions.



                    Let's notice that $3x cos x$ and $left( x^2 - 3 right) sin x$ have the same trigonometric argument. That means that the Harmonic Addition Theorem can be used: we can rewrite this as one trigonometric function.



                    HAT states that
                    begin{align}a cos x + b sin x &= A cos left(x + B right),
                    end{align}
                    where $a^2 + b^2 = A^2$ and $-frac{b}a = tan B$.



                    So, substituting $3x cos x$ for $a$ and $x^2 - 3$ for $b$, we find that $A^2 = x^4 + 3x^2 + 9$ and $frac{3 - x^2}{3x} = tan B$.



                    This of course has infinitely many solutions, although the factor of $A$ only contributes some complex roots. Ignoring questions of signs and branch choice, let's declare that $A = sqrt{x^4 + 3x^2 + 9}$ and $B = tan^{-1} frac{3 - x^2}{3x}$, or in other words
                    $$3x cos x + (x^2 - 3) sin x = sqrt{x^4 + 3x^2 + 9} cos left( x + tan^{-1} frac{3 - x^2}{3x} right).$$



                    The problem can therefore be rewritten as finding the zeroes of $sqrt{x^4 + 3x^2 + 9}$ and $cos left( x + tan^{-1} frac{3 - x^2}{3x} right)$, although I doubt you'll find any good closed form for the latter.



                    This WolframAlpha plot looks like it agrees with your image.






                    share|cite|improve this answer









                    $endgroup$



                    There's probably no closed-form solution for the real roots you're looking for, although you have a chance with the complex solutions.



                    Let's notice that $3x cos x$ and $left( x^2 - 3 right) sin x$ have the same trigonometric argument. That means that the Harmonic Addition Theorem can be used: we can rewrite this as one trigonometric function.



                    HAT states that
                    begin{align}a cos x + b sin x &= A cos left(x + B right),
                    end{align}
                    where $a^2 + b^2 = A^2$ and $-frac{b}a = tan B$.



                    So, substituting $3x cos x$ for $a$ and $x^2 - 3$ for $b$, we find that $A^2 = x^4 + 3x^2 + 9$ and $frac{3 - x^2}{3x} = tan B$.



                    This of course has infinitely many solutions, although the factor of $A$ only contributes some complex roots. Ignoring questions of signs and branch choice, let's declare that $A = sqrt{x^4 + 3x^2 + 9}$ and $B = tan^{-1} frac{3 - x^2}{3x}$, or in other words
                    $$3x cos x + (x^2 - 3) sin x = sqrt{x^4 + 3x^2 + 9} cos left( x + tan^{-1} frac{3 - x^2}{3x} right).$$



                    The problem can therefore be rewritten as finding the zeroes of $sqrt{x^4 + 3x^2 + 9}$ and $cos left( x + tan^{-1} frac{3 - x^2}{3x} right)$, although I doubt you'll find any good closed form for the latter.



                    This WolframAlpha plot looks like it agrees with your image.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 8 at 1:53









                    BladewoodBladewood

                    335213




                    335213












                    • $begingroup$
                      Thanks for such a nice manipulation. But still the expression $sqrt(x^4+3x^2+9)=0$ doesn't have the reql solutions. I will try to work with other factor
                      $endgroup$
                      – ersh
                      Jan 8 at 5:05


















                    • $begingroup$
                      Thanks for such a nice manipulation. But still the expression $sqrt(x^4+3x^2+9)=0$ doesn't have the reql solutions. I will try to work with other factor
                      $endgroup$
                      – ersh
                      Jan 8 at 5:05
















                    $begingroup$
                    Thanks for such a nice manipulation. But still the expression $sqrt(x^4+3x^2+9)=0$ doesn't have the reql solutions. I will try to work with other factor
                    $endgroup$
                    – ersh
                    Jan 8 at 5:05




                    $begingroup$
                    Thanks for such a nice manipulation. But still the expression $sqrt(x^4+3x^2+9)=0$ doesn't have the reql solutions. I will try to work with other factor
                    $endgroup$
                    – ersh
                    Jan 8 at 5:05


















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