The Math.SE question *$2\cos(2x) - 2\sin(x) = 0$* has attracted several
answers from high-rep users.

I am expanding @rhombic's comment into an answer.

\[ \begin{aligned} 2\cos(2x)-2\sin(x)&=0 \\ 2 - 4\sin^2(x)-2\sin(x)&=0 \\ 2\sin^2(x)+\sin(x) - 1&=0 \\ (2 \sin(x) - 1)(\sin(x) +1) &= 0 \\ \sin(x) = \frac12 \text{ or } \sin(x) &= -1 \\ x = \frac{\pi}{6}, \frac{5\pi}{6} \text{ (rejected) or } & \frac{3\pi}{2} \text{ (rejected)} \end{aligned} \]