giúp mình với : Sin^2 2x = cos^2x

`#kehuydiet`

$\sin^22x = \cos ^2 x$

\(⇔\left[ \begin{array}{l}\sin 2x=\cos x\\ -\sin 2x=\cos x\end{array} \right.\) 

\(⇔\left[ \begin{array}{l}x-\dfrac{\pi}4=x + k2\pi \\ x-\dfrac{\pi}4=-x + k2\pi \\ \dfrac{\pi}{2}+2x=x+k2\pi \\ \dfrac{\pi}{2}+2x=-x+k2\pi\end{array} \right.\) 

\(⇔\left[ \begin{array}{l} x= \dfrac{\pi}8 + k\pi \\ x=-\dfrac{\pi}{2}+k2\pi \\ x=-\dfrac{\pi}{6}+\dfrac{k2\pi}3\end{array} \right. (k\in \mathbb{Z})\)