Question

sin4x-sin2x-sinx=0

Answer 1

I would say:
\[\sin(4x)-\sin(2x)-\sin(x)=\Im \left[ e^{4ix}-e^{2ix}-e^{ix}\right]; \psi=e^{ix} \implies \]
\[\implies \psi^4-\psi^2-\psi=0 \implies \psi(\psi^3-\psi-1)=0\]
Then solve for Psi, then Pis=e^(ix) solve for x taking the imaginary part. Then you have solutions but factoring psi^3-psi-1=0 gives crappyyyy numbers