Đáp án:
$\left[\begin{array}{l}x =\arccos(-0,64620) + k2\pi\\x = \arccos0,19865 + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\dfrac{1}{\cos x} = 4\sin^2x + 6\cos x\qquad (*)$
$ĐKXĐ:\, x \ne \dfrac{\pi}{2} + n\pi\quad (n\in\Bbb Z)$
$(*)\Leftrightarrow 4(1 - \cos^2x)\cos x + 6\cos^2x - 1 = 0$
$\Leftrightarrow 4\cos^3x - 6\cos^2x - 4\cos x + 1 = 0$
$\Leftrightarrow \left[\begin{array}{l}\cos x \approx -0,64620\\\cos x \approx 0,19865\\\cos x \approx 1,9476\quad (loại)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x =\arccos(-0,64620) + k2\pi\\x = \arccos0,19865 + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
