giải phương trình: 3cos4x-sin^22x+cos2x-2=0

Đáp án:

$\left[\begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\x = \pm \dfrac{1}{2}\arccos\dfrac{6}{7}+ k\pi\end{array}\right.\quad (k \in \Bbb Z)$

Giải thích các bước giải:

$3\cos4x – \sin^22x + \cos2x – 2 = 0$

$\Leftrightarrow 3(2\cos^22x – 1) – (1-\cos^22x) + \cos2x – 2 = 0$

$\Leftrightarrow 7\cos^22x + \cos2x – 6 = 0$

$\Leftrightarrow \left[\begin{array}{l}\cos2x = -1\\\cos2x = \dfrac{6}{7}\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}2x = \pi + k2\pi\\2x = \pm \arccos\dfrac{6}{7}+ k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\x = \pm \dfrac{1}{2}\arccos\dfrac{6}{7}+ k\pi\end{array}\right.\quad (k \in \Bbb Z)$