Solution
= Solution
Use $2\cos^2x=1+\cos2x$ to reduce the equation to $\sin2x=\cos2x$. Equivalently, $\sqrt2\sin(2x-\pi/4)=0$, so $x=\pi/8+k\pi/2$ for an integer $k$. Restricting to $[0,\pi]$ gives
$$
\boxed{x=\frac\pi8,\qquad x=\frac{5\pi}{8}.}
$$