2cos⁶x + sin⁴x + cos2x =0

Đáp án:

$x = \dfrac{\pi}{2} + k\pi\quad (k\in \Bbb Z)$

Giải thích các bước giải:

$2\cos^6x + \sin^4x + \cos2x = 0$

$\Leftrightarrow 2\cos^6x + \cos^2x - \sin^2x + \sin^4x = 0$

$\Leftrightarrow \cos^2x(2\cos^4x + 1) - \sin^2x(1 - \sin^2x) = 0$

$\Leftrightarrow \cos^2x(2\cos^4x + 1) - \sin^2x\cos^2x = 0$

$\Leftrightarrow \cos^2x(2\cos^4x + 1 - \sin^2x) = 0$

$\Leftrightarrow \cos^2x(2\cos^4x + \cos^2x) = 0$

$\Leftrightarrow \cos^4x(2\cos^2x + 1) = 0$

$\Leftrightarrow \cos x = 0$

$\Leftrightarrow x = \dfrac{\pi}{2} + k\pi\quad (k\in \Bbb Z)$