Answer
Three points:
$(\dfrac{\pi}{6}, \dfrac{3\sqrt{3}}{2})$, $(\dfrac{5\pi}{6}, -\dfrac{3\sqrt{3}}{2})$ and $(\dfrac{3\pi}{2}, 0)$
Work Step by Step
$f(x)=g(x)+h(x)\rightarrow g(x)=2\cos{x}$ ; $h(x)=\sin{2x}$
$g'(x)=-2\sin{x}$; $h'(x)=2\cos{2x}$
$f'(x)=g'(x)+h'(x)=2\cos{2x}-2\sin{x}$
$f'(x)=0\rightarrow 2\cos{2x}-2\sin{x}=0$
To solve the trigonometric equation you can graph it (quick and efficient but not always possible) or algebraically as follows:
$2(1-2\sin^2{x})-2\sin{x}=0\rightarrow$
$2\sin^2{x}+\sin{x}-1=0\rightarrow \sin{x}=\dfrac{1}{2}$ or $\sin{x}=-1\rightarrow$
$x=\dfrac{\pi}{6}$, $x=\dfrac{5\pi}{6}$ or $\dfrac{3\pi}{2}.$
By plugging each value into the original function we get three points in the specified domain:
$(\dfrac{\pi}{6}, \dfrac{3\sqrt{3}}{2})$, $(\dfrac{5\pi}{6}, -\dfrac{3\sqrt{3}}{2})$ and $(\dfrac{3\pi}{2}, 0)$
