Giải thích các bước giải:
$\cos^2x - \sin2x = \sqrt{2} + \cos^2\left( \dfrac{\pi}{2} + x \right)$
`<=>`$\cos^2x - \sin2x = \sqrt{2} + \sin^2x$
$\Leftrightarrow 2\cos^2x - 2\sin2x = 2\sqrt{2} + 2\sin^2x$
$\Leftrightarrow 1 + \cos2x - 2\sin2x = 2\sqrt{2} + 1 - \cos2x$
$\Leftrightarrow \cos2x - \sin2x = \sqrt{2}$
$\Leftrightarrow \cos \left( 2x - \dfrac{\pi}{4} \right) = 1$
$\Leftrightarrow \cos \left( 2x - \dfrac{\pi}{4} \right) = \cos 0$
`<=>`$2x - \dfrac{\pi}{4} = 2k\pi$
$\Leftrightarrow x = \dfrac{\pi}{8} + k\pi$
