We first use the double angle formula to write \(\sin (2x)\) in terms of trig functions of \(x\) alone.
\begin{align*}
\sin (2x) -\cos (x) \amp = 0\\
2\sin (x) \cos (x) -\cos (x) \amp = 0
\end{align*}
Once we have all the trig functions in terms of a single angle, we try to write the equation in terms of a single trig function. In this case, we can factor the left side to separate the trig functions.
\begin{align*}
2\sin (x) \cos (x) -\cos (x) \amp = 0\\
\cos (x) (2\sin (x) - 1) \amp = 0 \amp\amp \blert{\text{Set each factor equal to zero.}}\\
\cos (x) = 0 \qquad 2\sin (x) - 1 \amp = 0 \amp\amp \blert{\text{Solve each equation.}}\\
\sin (x) \amp = \dfrac{1}{2}\\
x = \dfrac{\pi}{2},~\dfrac{3\pi}{2} \qquad x = \dfrac{\pi}{6}\amp,~\dfrac{5\pi}{6}
\end{align*}
There are four solutions, \(x = \dfrac{\pi}{2},~\dfrac{3\pi}{2},~\dfrac{\pi}{6},\) and \(\dfrac{5\pi}{6}\text{.}\)