Solution
= Solution
<Euler's formula> gives
$$
\sin x=\frac{e^{ix}-e^{-ix}}{2i},\qquad\cos x=\frac{e^{ix}+e^{-ix}}2.
$$
Cubing the expression for the <sine>, and grouping the $e^{3ix}$ and $e^{-3ix}$ terms, gives
$$
\sin^3x=-\frac14\bigl(\sin3x-3\sin x\bigr),\qquad\sin3x=3\sin x-4\sin^3x.
$$
The required equation therefore becomes $\sin x(1-4\sin^2x)=0$. The complete real solution set is
$$
\boxed{x=n\pi\quad\text{or}\quad x=n\pi\pm\frac\pi6,\qquad n\in\mathbb Z.}
$$