Đáp án:
1) $x = \pm \dfrac{2\pi}{3}+k2\pi\quad (k \in \Bbb Z)$
2) $x = -\dfrac{\pi}{2} + k2\pi\quad (k\in \Bbb Z)$
Giải thích các bước giải:
1) $\cos^4x - \sin^4x + 5 \cos x + 3 = 0$
$\Leftrightarrow (\cos^2x - \sin^2x)(\cos^2x + \sin^2x) + 5 \cos x + 3 = 0$
$\Leftrightarrow 1.(2\cos^2x - 1)+ 5 \cos x + 3 = 0$
$\Leftrightarrow 2\cos^2x + 5\cos x + 2 = 0$
$\Leftrightarrow \left[\begin{array}{l}\cos x = -\dfrac{1}{2}\quad (nhận)\\\cos x = -2\quad (loại)\end{array}\right.$
$\Leftrightarrow x = \pm \dfrac{2\pi}{3}+k2\pi\quad (k \in \Bbb Z)$
2) $3\cos^2x + \cos^2x.\sin x = 8(1+\sin x)$
$\Leftrightarrow 3(1 - \sin^2x) + (1-\sin^2x)\sin x = 8 + 8\sin x$
$\Leftrightarrow \sin^3x + 3\sin^2x + 7\sin x + 5 = 0$
$\Leftrightarrow \sin x = -1$
$\Leftrightarrow x = -\dfrac{\pi}{2} + k2\pi\quad (k\in \Bbb Z)$
