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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., $\phi$,

$$\sin(2k+\phi)\sin(3k+\phi)+\alpha\sin^2(2k+\phi)+\beta\sin^2k\sin(3k+\phi)\csc(2k+\phi)+A\sin^2k=0$$

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

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

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  • $\begingroup$ Maybe try the product to sum formulas? $\endgroup$ – YiFan Sep 6 '18 at 0:17
  • $\begingroup$ Its I think only possible to solve it numerically. $\endgroup$ – AtoZ Sep 7 '18 at 3:18

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