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?
FullSimplify@Solve[-Sin[3*k + φ] + α* Sin[2*k + φ] + β*(Sin^2)[k]* Sin[2*k + φ] == 0, φ]
(wrapped inQuiet
if you want to suppress the warning message) does give a reasonably simplified result. $\endgroup$Solve[-Sin[3*k+φ]+α*Sin[2*k+φ]+β*(Sin^2)[k]*Sin[2*k+φ]==0//TrigExpand,φ]//Simplify
orSolve[-Sin[3*k+φ]+α*Sin[2*k+φ]+β*(Sin^2)[k]*Sin[2*k+φ]==0/.φ->ArcTan[x]//TrigExpand,x]//Simplify
$\endgroup$Assumptions -> {-Pi <= k <= Pi}
inFullSimplify
but, in this case, it does not change the result. $\endgroup$