Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{6} + k\dfrac{2\pi}{3}\\x = \dfrac{\pi}{18} + k\dfrac{2\pi}{3}\\x = \dfrac{5\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in \Bbb Z)$
Giải thích các bước giải:
$\cos^2\left(3x + \dfrac{\pi}{2}\right) - \cos^23x - 3\cos\left(\dfrac{\pi}{2} - 3x\right) + 2 = 0$
$\Leftrightarrow (-\sin3x)^2 - (1-\sin^23x) - 3\sin3x + 2 = 0$
$\Leftrightarrow 2\sin^23x - 3\sin3x + 1 = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin3x = 1\\\sin3x = \dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}3x = \dfrac{\pi}{2} + k2\pi\\3x = \dfrac{\pi}{6} + k2\pi\\3x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} + k\dfrac{2\pi}{3}\\x = \dfrac{\pi}{18} + k\dfrac{2\pi}{3}\\x = \dfrac{5\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in \Bbb Z)$
