Answer:
[tex]\huge\boxed{x\in\left\{\dfrac{\pi}{6};\ \dfrac{\pi}{2}\ \dfrac{3\pi}{2};\ \dfrac{5\pi}{6}\right\}}[/tex]
Step-by-step explanation:
[tex]\cos x=\sin2x\qquad|\text{use}\ \sin\theta=2\sin\theta\cos\theta\\\\\cos x=2\sin x\cos x\qquad|\text{subtract}\ \cos x\ \text{from both sides}\\\\0=2\sin x\cos x-\cos x\\\\2\sin x\cos x-\cos x=0\qquad|\text{distribute}\\\\\cos x(2\sin x-1)=0\iff\underbrace{\cos x=0}_{(1)}\ \vee\ \underbrace{2\sin x-1=0}_{(2)}[/tex]
[tex](1)\\\cos x=0\Rightarrow x=\dfrac{\pi}{2}+k\pi;\ k\in\mathbb{Z}[/tex]
[tex](2)\\2\sin x-1=0\qquad|\text{add 1 to both sides}\\\\2\sin x=1\qquad|\text{divide sides by 2}\\\\\sin x=\dfrac{1}{2}\Rightarrow x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi[/tex]
[tex]\text{From}\ (1)\ \text{and}\ (2)\ \text{we have}\\\\x=\dfrac{\pi}{2}+k\pi\ \vee\ x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi\\\\\text{We have the interval}\ x\in[0;2\pi).\ \text{Therefore the solution is:}\\\\x\in\left\{\dfrac{\pi}{6};\ \dfrac{\pi}{2}\ \dfrac{3\pi}{2};\ \dfrac{5\pi}{6}\right\}[/tex]
