Đáp án:
$x \in \left\{\dfrac{\pi}{4};\dfrac{\pi}{2};\dfrac{3\pi}{4};\pi;\dfrac{3\pi}{2};\dfrac{5\pi}{4};\dfrac{7\pi}{4}\right\}$
Giải thích các bước giải:
$|f(\cos 2x)|=1\\ \Leftrightarrow f(\cos 2x)= \pm 1$
Dựa vào BBT, $f(\cos 2x)= \pm 1$
$\Rightarrow \left\{\begin{array}{l} \cos 2x =-1 \\ \cos 2x =1 \\ \cos 2x =0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 2x =\pi + k 2 \pi ( k \in \mathbb{Z}) \\ 2x = k 2 \pi ( k \in \mathbb{Z})\\ 2x =\dfrac{\pi}{2} + k \pi( k \in \mathbb{Z}) \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x =\dfrac{\pi}{2} + k \pi ( k \in \mathbb{Z}) \\ x = k \pi ( k \in \mathbb{Z})\\ x =\dfrac{\pi}{4} + \dfrac{k \pi}{2}( k \in \mathbb{Z}) \end{array} \right.\\ x \in (0;2\pi ) \Rightarrow x \in \left\{\dfrac{\pi}{4};\dfrac{\pi}{2};\dfrac{3\pi}{4};\pi;\dfrac{3\pi}{2};\dfrac{5\pi}{4};\dfrac{7\pi}{4}\right\}$
