Using the R method for finding all solutions to $sin(2a) - cos(2a) = frac{sqrt{6}}{2}$. My solution differs...












3












$begingroup$



How many solutions does
$$sin(2a) - cos(2a) = frac{sqrt{6}}{2}$$
have between $-90^circ$ and $90^circ$?




I used the R method and got
$$2a-45^circ = arcsinleft(frac{sqrt{3}}{2}right).$$
Since $a$ is between $-90^circ$ and $90^circ$, then $2a$ is between $-180^circ$ and $180^circ$. The RHS can be $60^circ$, $120^circ$, $-240^circ$, and $-300^circ$. Only $60^circ$ and $120^circ$ fit the criteria, but the answer is 4 solutions.




Where did I go wrong?







trigonometry







share|cite|improve this question











$endgroup$

















    3












    $begingroup$



    How many solutions does
    $$sin(2a) - cos(2a) = frac{sqrt{6}}{2}$$
    have between $-90^circ$ and $90^circ$?




    I used the R method and got
    $$2a-45^circ = arcsinleft(frac{sqrt{3}}{2}right).$$
    Since $a$ is between $-90^circ$ and $90^circ$, then $2a$ is between $-180^circ$ and $180^circ$. The RHS can be $60^circ$, $120^circ$, $-240^circ$, and $-300^circ$. Only $60^circ$ and $120^circ$ fit the criteria, but the answer is 4 solutions.




    Where did I go wrong?







    trigonometry







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$



      How many solutions does
      $$sin(2a) - cos(2a) = frac{sqrt{6}}{2}$$
      have between $-90^circ$ and $90^circ$?




      I used the R method and got
      $$2a-45^circ = arcsinleft(frac{sqrt{3}}{2}right).$$
      Since $a$ is between $-90^circ$ and $90^circ$, then $2a$ is between $-180^circ$ and $180^circ$. The RHS can be $60^circ$, $120^circ$, $-240^circ$, and $-300^circ$. Only $60^circ$ and $120^circ$ fit the criteria, but the answer is 4 solutions.




      Where did I go wrong?







      trigonometry







      share|cite|improve this question











      $endgroup$





      How many solutions does
      $$sin(2a) - cos(2a) = frac{sqrt{6}}{2}$$
      have between $-90^circ$ and $90^circ$?




      I used the R method and got
      $$2a-45^circ = arcsinleft(frac{sqrt{3}}{2}right).$$
      Since $a$ is between $-90^circ$ and $90^circ$, then $2a$ is between $-180^circ$ and $180^circ$. The RHS can be $60^circ$, $120^circ$, $-240^circ$, and $-300^circ$. Only $60^circ$ and $120^circ$ fit the criteria, but the answer is 4 solutions.




      Where did I go wrong?







      trigonometry




      trigonometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 14 at 12:12









      Namaste

      1




      1










      asked Jan 14 at 2:32









      SuperMage1SuperMage1

      916211




      916211






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          You are right. Here is the image of the function using google.



          enter image description here






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, but i want to know whats wrong with this solution that got 4 as the answer. Sin 2a - cos 2a = root 6 /2, then he squared it, getting 1-2sin2acos2a= 3/2 which is sin 4a = -1/2 which does have 4 solutions.
            $endgroup$
            – SuperMage1
            Jan 14 at 2:54










          • $begingroup$
            Since I don't see how they got 4 solutions, there is no way to explain what they did wrong. Do you know which solutions they got?
            $endgroup$
            – Andrei
            Jan 14 at 2:56










          • $begingroup$
            No solutions were given
            $endgroup$
            – SuperMage1
            Jan 14 at 2:58










          • $begingroup$
            @SuperMage1 when we square we add solutions $x=2$ square $x^2 = 4$ two solutions -/+2. If you square you need to return to the original equation and check the answers
            $endgroup$
            – Ameryr
            Jan 14 at 3:25










          • $begingroup$
            @Ameryr That's why you have a plot of the original equation, so you can confirm that there are only two solutions, not 4
            $endgroup$
            – Andrei
            Jan 14 at 3:28



















          0












          $begingroup$

          Hint:



          $$2a-45^circ=180^circ n+(-1)^narcsindfrac{sqrt3}2$$



          If $n$ is odd$=2m+1$(say)



          $$2a-45=360m+180-60$$



          But $-180-45le2a-45le180-45$



          $-225le360m+120le135$



          $?le mle?$



          What if $n$ is even $=2m$(say)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry, but wat do i do after gettong the bounds in this solution, and it does seem a bit more tedious.
            $endgroup$
            – SuperMage1
            Jan 14 at 3:03










          • $begingroup$
            @SuperMage1, Please let me know if u find one better
            $endgroup$
            – lab bhattacharjee
            Jan 14 at 3:24



















          0












          $begingroup$

          $sin(2a-45) = frac{sqrt{3}}{2}$



          $2a-45 = 60 + 360n Rightarrow a = 52.5 + 180n$



          $2a-45 = 120+ 360n Rightarrow a = 82.5 + 180n$



          Either $a = 52.5 $ or $a = 82.5$. If you square



          $sin(4a) = frac{-1}{2}$



          $4a = 210 + 360n Rightarrow a = 52.5 + 90n $



          $4a = 330 + 360n Rightarrow a = 82.5 + 90n $
          First two solution
          $a = 52.5 $



          $a= 52.5 - 90 = -38.5$



          The other two
          $a = 82.5 , 82.5 - 90 = -7.5$



          In both cases the second solution is rejected see
          Checking $sin(-15) - cos(-15) = frac{-sqrt{6}}{2}$ which is wrong same for the other






          share|cite|improve this answer









          $endgroup$













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            You are right. Here is the image of the function using google.



            enter image description here






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks, but i want to know whats wrong with this solution that got 4 as the answer. Sin 2a - cos 2a = root 6 /2, then he squared it, getting 1-2sin2acos2a= 3/2 which is sin 4a = -1/2 which does have 4 solutions.
              $endgroup$
              – SuperMage1
              Jan 14 at 2:54










            • $begingroup$
              Since I don't see how they got 4 solutions, there is no way to explain what they did wrong. Do you know which solutions they got?
              $endgroup$
              – Andrei
              Jan 14 at 2:56










            • $begingroup$
              No solutions were given
              $endgroup$
              – SuperMage1
              Jan 14 at 2:58










            • $begingroup$
              @SuperMage1 when we square we add solutions $x=2$ square $x^2 = 4$ two solutions -/+2. If you square you need to return to the original equation and check the answers
              $endgroup$
              – Ameryr
              Jan 14 at 3:25










            • $begingroup$
              @Ameryr That's why you have a plot of the original equation, so you can confirm that there are only two solutions, not 4
              $endgroup$
              – Andrei
              Jan 14 at 3:28
















            1












            $begingroup$

            You are right. Here is the image of the function using google.



            enter image description here






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks, but i want to know whats wrong with this solution that got 4 as the answer. Sin 2a - cos 2a = root 6 /2, then he squared it, getting 1-2sin2acos2a= 3/2 which is sin 4a = -1/2 which does have 4 solutions.
              $endgroup$
              – SuperMage1
              Jan 14 at 2:54










            • $begingroup$
              Since I don't see how they got 4 solutions, there is no way to explain what they did wrong. Do you know which solutions they got?
              $endgroup$
              – Andrei
              Jan 14 at 2:56










            • $begingroup$
              No solutions were given
              $endgroup$
              – SuperMage1
              Jan 14 at 2:58










            • $begingroup$
              @SuperMage1 when we square we add solutions $x=2$ square $x^2 = 4$ two solutions -/+2. If you square you need to return to the original equation and check the answers
              $endgroup$
              – Ameryr
              Jan 14 at 3:25










            • $begingroup$
              @Ameryr That's why you have a plot of the original equation, so you can confirm that there are only two solutions, not 4
              $endgroup$
              – Andrei
              Jan 14 at 3:28














            1












            1








            1





            $begingroup$

            You are right. Here is the image of the function using google.



            enter image description here






            share|cite|improve this answer









            $endgroup$



            You are right. Here is the image of the function using google.



            enter image description here







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 14 at 2:47









            AndreiAndrei

            13.2k21230




            13.2k21230












            • $begingroup$
              Thanks, but i want to know whats wrong with this solution that got 4 as the answer. Sin 2a - cos 2a = root 6 /2, then he squared it, getting 1-2sin2acos2a= 3/2 which is sin 4a = -1/2 which does have 4 solutions.
              $endgroup$
              – SuperMage1
              Jan 14 at 2:54










            • $begingroup$
              Since I don't see how they got 4 solutions, there is no way to explain what they did wrong. Do you know which solutions they got?
              $endgroup$
              – Andrei
              Jan 14 at 2:56










            • $begingroup$
              No solutions were given
              $endgroup$
              – SuperMage1
              Jan 14 at 2:58










            • $begingroup$
              @SuperMage1 when we square we add solutions $x=2$ square $x^2 = 4$ two solutions -/+2. If you square you need to return to the original equation and check the answers
              $endgroup$
              – Ameryr
              Jan 14 at 3:25










            • $begingroup$
              @Ameryr That's why you have a plot of the original equation, so you can confirm that there are only two solutions, not 4
              $endgroup$
              – Andrei
              Jan 14 at 3:28


















            • $begingroup$
              Thanks, but i want to know whats wrong with this solution that got 4 as the answer. Sin 2a - cos 2a = root 6 /2, then he squared it, getting 1-2sin2acos2a= 3/2 which is sin 4a = -1/2 which does have 4 solutions.
              $endgroup$
              – SuperMage1
              Jan 14 at 2:54










            • $begingroup$
              Since I don't see how they got 4 solutions, there is no way to explain what they did wrong. Do you know which solutions they got?
              $endgroup$
              – Andrei
              Jan 14 at 2:56










            • $begingroup$
              No solutions were given
              $endgroup$
              – SuperMage1
              Jan 14 at 2:58










            • $begingroup$
              @SuperMage1 when we square we add solutions $x=2$ square $x^2 = 4$ two solutions -/+2. If you square you need to return to the original equation and check the answers
              $endgroup$
              – Ameryr
              Jan 14 at 3:25










            • $begingroup$
              @Ameryr That's why you have a plot of the original equation, so you can confirm that there are only two solutions, not 4
              $endgroup$
              – Andrei
              Jan 14 at 3:28
















            $begingroup$
            Thanks, but i want to know whats wrong with this solution that got 4 as the answer. Sin 2a - cos 2a = root 6 /2, then he squared it, getting 1-2sin2acos2a= 3/2 which is sin 4a = -1/2 which does have 4 solutions.
            $endgroup$
            – SuperMage1
            Jan 14 at 2:54




            $begingroup$
            Thanks, but i want to know whats wrong with this solution that got 4 as the answer. Sin 2a - cos 2a = root 6 /2, then he squared it, getting 1-2sin2acos2a= 3/2 which is sin 4a = -1/2 which does have 4 solutions.
            $endgroup$
            – SuperMage1
            Jan 14 at 2:54












            $begingroup$
            Since I don't see how they got 4 solutions, there is no way to explain what they did wrong. Do you know which solutions they got?
            $endgroup$
            – Andrei
            Jan 14 at 2:56




            $begingroup$
            Since I don't see how they got 4 solutions, there is no way to explain what they did wrong. Do you know which solutions they got?
            $endgroup$
            – Andrei
            Jan 14 at 2:56












            $begingroup$
            No solutions were given
            $endgroup$
            – SuperMage1
            Jan 14 at 2:58




            $begingroup$
            No solutions were given
            $endgroup$
            – SuperMage1
            Jan 14 at 2:58












            $begingroup$
            @SuperMage1 when we square we add solutions $x=2$ square $x^2 = 4$ two solutions -/+2. If you square you need to return to the original equation and check the answers
            $endgroup$
            – Ameryr
            Jan 14 at 3:25




            $begingroup$
            @SuperMage1 when we square we add solutions $x=2$ square $x^2 = 4$ two solutions -/+2. If you square you need to return to the original equation and check the answers
            $endgroup$
            – Ameryr
            Jan 14 at 3:25












            $begingroup$
            @Ameryr That's why you have a plot of the original equation, so you can confirm that there are only two solutions, not 4
            $endgroup$
            – Andrei
            Jan 14 at 3:28




            $begingroup$
            @Ameryr That's why you have a plot of the original equation, so you can confirm that there are only two solutions, not 4
            $endgroup$
            – Andrei
            Jan 14 at 3:28











            0












            $begingroup$

            Hint:



            $$2a-45^circ=180^circ n+(-1)^narcsindfrac{sqrt3}2$$



            If $n$ is odd$=2m+1$(say)



            $$2a-45=360m+180-60$$



            But $-180-45le2a-45le180-45$



            $-225le360m+120le135$



            $?le mle?$



            What if $n$ is even $=2m$(say)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Sorry, but wat do i do after gettong the bounds in this solution, and it does seem a bit more tedious.
              $endgroup$
              – SuperMage1
              Jan 14 at 3:03










            • $begingroup$
              @SuperMage1, Please let me know if u find one better
              $endgroup$
              – lab bhattacharjee
              Jan 14 at 3:24
















            0












            $begingroup$

            Hint:



            $$2a-45^circ=180^circ n+(-1)^narcsindfrac{sqrt3}2$$



            If $n$ is odd$=2m+1$(say)



            $$2a-45=360m+180-60$$



            But $-180-45le2a-45le180-45$



            $-225le360m+120le135$



            $?le mle?$



            What if $n$ is even $=2m$(say)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Sorry, but wat do i do after gettong the bounds in this solution, and it does seem a bit more tedious.
              $endgroup$
              – SuperMage1
              Jan 14 at 3:03










            • $begingroup$
              @SuperMage1, Please let me know if u find one better
              $endgroup$
              – lab bhattacharjee
              Jan 14 at 3:24














            0












            0








            0





            $begingroup$

            Hint:



            $$2a-45^circ=180^circ n+(-1)^narcsindfrac{sqrt3}2$$



            If $n$ is odd$=2m+1$(say)



            $$2a-45=360m+180-60$$



            But $-180-45le2a-45le180-45$



            $-225le360m+120le135$



            $?le mle?$



            What if $n$ is even $=2m$(say)






            share|cite|improve this answer









            $endgroup$



            Hint:



            $$2a-45^circ=180^circ n+(-1)^narcsindfrac{sqrt3}2$$



            If $n$ is odd$=2m+1$(say)



            $$2a-45=360m+180-60$$



            But $-180-45le2a-45le180-45$



            $-225le360m+120le135$



            $?le mle?$



            What if $n$ is even $=2m$(say)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 14 at 2:55









            lab bhattacharjeelab bhattacharjee

            227k15158276




            227k15158276












            • $begingroup$
              Sorry, but wat do i do after gettong the bounds in this solution, and it does seem a bit more tedious.
              $endgroup$
              – SuperMage1
              Jan 14 at 3:03










            • $begingroup$
              @SuperMage1, Please let me know if u find one better
              $endgroup$
              – lab bhattacharjee
              Jan 14 at 3:24


















            • $begingroup$
              Sorry, but wat do i do after gettong the bounds in this solution, and it does seem a bit more tedious.
              $endgroup$
              – SuperMage1
              Jan 14 at 3:03










            • $begingroup$
              @SuperMage1, Please let me know if u find one better
              $endgroup$
              – lab bhattacharjee
              Jan 14 at 3:24
















            $begingroup$
            Sorry, but wat do i do after gettong the bounds in this solution, and it does seem a bit more tedious.
            $endgroup$
            – SuperMage1
            Jan 14 at 3:03




            $begingroup$
            Sorry, but wat do i do after gettong the bounds in this solution, and it does seem a bit more tedious.
            $endgroup$
            – SuperMage1
            Jan 14 at 3:03












            $begingroup$
            @SuperMage1, Please let me know if u find one better
            $endgroup$
            – lab bhattacharjee
            Jan 14 at 3:24




            $begingroup$
            @SuperMage1, Please let me know if u find one better
            $endgroup$
            – lab bhattacharjee
            Jan 14 at 3:24











            0












            $begingroup$

            $sin(2a-45) = frac{sqrt{3}}{2}$



            $2a-45 = 60 + 360n Rightarrow a = 52.5 + 180n$



            $2a-45 = 120+ 360n Rightarrow a = 82.5 + 180n$



            Either $a = 52.5 $ or $a = 82.5$. If you square



            $sin(4a) = frac{-1}{2}$



            $4a = 210 + 360n Rightarrow a = 52.5 + 90n $



            $4a = 330 + 360n Rightarrow a = 82.5 + 90n $
            First two solution
            $a = 52.5 $



            $a= 52.5 - 90 = -38.5$



            The other two
            $a = 82.5 , 82.5 - 90 = -7.5$



            In both cases the second solution is rejected see
            Checking $sin(-15) - cos(-15) = frac{-sqrt{6}}{2}$ which is wrong same for the other






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              $sin(2a-45) = frac{sqrt{3}}{2}$



              $2a-45 = 60 + 360n Rightarrow a = 52.5 + 180n$



              $2a-45 = 120+ 360n Rightarrow a = 82.5 + 180n$



              Either $a = 52.5 $ or $a = 82.5$. If you square



              $sin(4a) = frac{-1}{2}$



              $4a = 210 + 360n Rightarrow a = 52.5 + 90n $



              $4a = 330 + 360n Rightarrow a = 82.5 + 90n $
              First two solution
              $a = 52.5 $



              $a= 52.5 - 90 = -38.5$



              The other two
              $a = 82.5 , 82.5 - 90 = -7.5$



              In both cases the second solution is rejected see
              Checking $sin(-15) - cos(-15) = frac{-sqrt{6}}{2}$ which is wrong same for the other






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                $sin(2a-45) = frac{sqrt{3}}{2}$



                $2a-45 = 60 + 360n Rightarrow a = 52.5 + 180n$



                $2a-45 = 120+ 360n Rightarrow a = 82.5 + 180n$



                Either $a = 52.5 $ or $a = 82.5$. If you square



                $sin(4a) = frac{-1}{2}$



                $4a = 210 + 360n Rightarrow a = 52.5 + 90n $



                $4a = 330 + 360n Rightarrow a = 82.5 + 90n $
                First two solution
                $a = 52.5 $



                $a= 52.5 - 90 = -38.5$



                The other two
                $a = 82.5 , 82.5 - 90 = -7.5$



                In both cases the second solution is rejected see
                Checking $sin(-15) - cos(-15) = frac{-sqrt{6}}{2}$ which is wrong same for the other






                share|cite|improve this answer









                $endgroup$



                $sin(2a-45) = frac{sqrt{3}}{2}$



                $2a-45 = 60 + 360n Rightarrow a = 52.5 + 180n$



                $2a-45 = 120+ 360n Rightarrow a = 82.5 + 180n$



                Either $a = 52.5 $ or $a = 82.5$. If you square



                $sin(4a) = frac{-1}{2}$



                $4a = 210 + 360n Rightarrow a = 52.5 + 90n $



                $4a = 330 + 360n Rightarrow a = 82.5 + 90n $
                First two solution
                $a = 52.5 $



                $a= 52.5 - 90 = -38.5$



                The other two
                $a = 82.5 , 82.5 - 90 = -7.5$



                In both cases the second solution is rejected see
                Checking $sin(-15) - cos(-15) = frac{-sqrt{6}}{2}$ which is wrong same for the other







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 14 at 3:33









                AmeryrAmeryr

                685311




                685311






























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                    Human spaceflight

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                    Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
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                    Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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                    0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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                    張江高科駅

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                    張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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