How to solve complicated equation for an argument of sine

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I have an equation which was quite complicated in the begining and I have simplified it upto this point with multiple sine functions and I want to solve for an argument of sine i.e., $phi$,



$$sin(2k+phi)sin(3k+phi)+alphasin^2(2k+phi)+betasin^2ksin(3k+phi)csc(2k+phi)+Asin^2k=0$$



$alpha$, $beta$, $A$ are constants and $k$ is a variable (a number between $-pi$ to $pi$).



I guess if I try to use $sin(a+b)=sin acos b+cos asin b$ at this point, it becomes even more complicated, so is there any alternate trick to solve it for $phi$?




calculus real-analysis linear-algebra trigonometry




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    I have an equation which was quite complicated in the begining and I have simplified it upto this point with multiple sine functions and I want to solve for an argument of sine i.e., $phi$,



    $$sin(2k+phi)sin(3k+phi)+alphasin^2(2k+phi)+betasin^2ksin(3k+phi)csc(2k+phi)+Asin^2k=0$$



    $alpha$, $beta$, $A$ are constants and $k$ is a variable (a number between $-pi$ to $pi$).



    I guess if I try to use $sin(a+b)=sin acos b+cos asin b$ at this point, it becomes even more complicated, so is there any alternate trick to solve it for $phi$?




    calculus real-analysis linear-algebra trigonometry




    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I have an equation which was quite complicated in the begining and I have simplified it upto this point with multiple sine functions and I want to solve for an argument of sine i.e., $phi$,



      $$sin(2k+phi)sin(3k+phi)+alphasin^2(2k+phi)+betasin^2ksin(3k+phi)csc(2k+phi)+Asin^2k=0$$



      $alpha$, $beta$, $A$ are constants and $k$ is a variable (a number between $-pi$ to $pi$).



      I guess if I try to use $sin(a+b)=sin acos b+cos asin b$ at this point, it becomes even more complicated, so is there any alternate trick to solve it for $phi$?




      calculus real-analysis linear-algebra trigonometry




      share|cite|improve this question














      I have an equation which was quite complicated in the begining and I have simplified it upto this point with multiple sine functions and I want to solve for an argument of sine i.e., $phi$,



      $$sin(2k+phi)sin(3k+phi)+alphasin^2(2k+phi)+betasin^2ksin(3k+phi)csc(2k+phi)+Asin^2k=0$$



      $alpha$, $beta$, $A$ are constants and $k$ is a variable (a number between $-pi$ to $pi$).



      I guess if I try to use $sin(a+b)=sin acos b+cos asin b$ at this point, it becomes even more complicated, so is there any alternate trick to solve it for $phi$?





      calculus real-analysis linear-algebra trigonometry





      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 23 at 4:21

























      asked Aug 23 at 4:06









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