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I want to solve a trigonometric equation involving a few sine functions, for an argument of sine function. The equation in its simplest form is $$-\sin(3k+\phi)+\alpha\sin(2k+\phi)+\beta\sin^2k\sin(2k+\phi)=0$$, where $\alpha$ and $\beta$ constants and $k$ is a variable (a number between $-\pi$ to $\pi$), and I want to solve it for $\phi$. I tried with Solve

 Solve[-Sin[3*k + φ] + α*
 Sin[2*k + φ] + β*(Sin^2)[k]*
 Sin[2*k + φ] == 0, φ]

I get a long output, also with an error message

Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

How do I get a correct output? which is simplified a bit?

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    $\begingroup$ FullSimplify@Solve[-Sin[3*k + φ] + α* Sin[2*k + φ] + β*(Sin^2)[k]* Sin[2*k + φ] == 0, φ] (wrapped in Quiet if you want to suppress the warning message) does give a reasonably simplified result. $\endgroup$ – kglr Aug 23 '18 at 6:02
  • $\begingroup$ Thanks. How do I put in the condition on $k$ that $-\pi\leq k\leq\pi$? It may simplify the answer a bit further? $\endgroup$ – AtoZ Aug 23 '18 at 6:12
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    $\begingroup$ Solve[-Sin[3*k+φ]+α*Sin[2*k+φ]+β*(Sin^2)[k]*Sin[2*k+φ]==0//TrigExpand,φ]//Simplify or Solve[-Sin[3*k+φ]+α*Sin[2*k+φ]+β*(Sin^2)[k]*Sin[2*k+φ]==0/.φ->ArcTan[x]//TrigExpand,x]//Simplify $\endgroup$ – matrix89 Aug 23 '18 at 6:15
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    $\begingroup$ AtoZ, you can use the option Assumptions -> {-Pi <= k <= Pi} in FullSimplify but, in this case, it does not change the result. $\endgroup$ – kglr Aug 23 '18 at 6:21
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If you use the "tangent half angle substitution" you force the constraint -Pi<k <Pi, -Pi<\[CurlyPhi]<Pi

eq = (-Sin[3*k + \[CurlyPhi]] + \[Alpha]*Sin[2*k + \[CurlyPhi]] + \[Beta]*(Sin^2)[k]*Sin[2*k + \[CurlyPhi]] 
/. {k -> 2 ArcTan[uk], \[CurlyPhi] ->2 ArcTan[u\[CurlyPhi]]} // TrigExpand  ) /.  (Sin^2)[2 ArcTan[uk]] -> (TrigExpand[Sin[2 ArcTan[uk]]])^2 // Simplify

Now the equation is rational in uk, u\[CurlyPhi] and can be solved without warning and ConditionalExpression

Solve[eq == 0 , u\[CurlyPhi]] /. {u\[CurlyPhi] -> Tan[\[CurlyPhi]/2],uk -> Tan[k/2]} //Simplify

enter image description here

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  • $\begingroup$ Thanks. There is a twin equation I am trying to solve which seems slightly harder Solve[-Sin[3*k + \[CurlyPhi]]/ Sin[2*k + \[CurlyPhi]] + \[Alpha] + (\[Mu]*(Sin^2)[ k])/((Sin^2)[2*k + \[CurlyPhi]] + \[Beta]*(Sin^2)[k]) == 0, \[CurlyPhi]], solve is unable to handle it. Any suggestion on this one? $\endgroup$ – AtoZ Aug 23 '18 at 9:58
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    $\begingroup$ My solution with this new equation works, but the solution contains Root-objects. $\endgroup$ – Ulrich Neumann Aug 23 '18 at 10:05
  • $\begingroup$ Root objects? Do we get the solution(s) \[CurlyPhi] somehow? Can you please show the code so I can put it and try to run. Thanks $\endgroup$ – AtoZ Aug 23 '18 at 10:17
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    $\begingroup$ (-Sin[3*k + \[CurlyPhi]]/ Sin[2*k + \[CurlyPhi]] + \[Alpha] + (\[Mu]*Sin[k]^2)/( Sin [2*k + \[CurlyPhi]]^2 + \[Beta]*Sin[k]^2) /. {k -> 2 ArcTan[uk], \[CurlyPhi] -> 2 ArcTan[u\[CurlyPhi]]} // TrigExpand ) /. (Sin^2)[ 2 ArcTan[uk]] -> (TrigExpand[Sin[2 ArcTan[uk]]])^2 // Simplify Solve[% == 0, u\[CurlyPhi]] $\endgroup$ – Ulrich Neumann Aug 23 '18 at 10:20
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    $\begingroup$ What means "truncated solution"? The analytical solution can only be simplified for given parametervalues \[Alpha], \[Beta] and/or additional information about the parameters. $\endgroup$ – Ulrich Neumann Aug 23 '18 at 10:41

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