Đáp án:
$\begin{array}{l}a) \,\left[\begin{array}{l}x = k\pi\\x = -\dfrac{\pi}{6} + k2\pi\\x = \dfrac{7\pi}{6} +k2\pi\end{array}\right. \quad (k \in \Bbb Z)\\b)\,\left[\begin{array}{l}x =\dfrac{5\pi}{12} +k2\pi\\x = \dfrac{11\pi}{12} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\end{array}$
Giải thích các bước giải:
$\begin{array}{l}a)\,\,\cos2x - \sin x = 1\\ \Leftrightarrow 1 - 2\sin^2x - \sin x = 1\\ \Leftrightarrow 2\sin^2x + \sin x = 0\\ \Leftrightarrow \sin x(2\sin x + 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin x =0\\\sin x = - \dfrac{1}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\x = -\dfrac{\pi}{6} + k2\pi\\x = \dfrac{7\pi}{6} +k2\pi\end{array}\right. \quad (k \in \Bbb Z)\\ b)\,\,\sqrt3\sin x - \cos x = \sqrt2\\ \Leftrightarrow \dfrac{\sqrt3}{2}\sin x - \dfrac{1}{2}\cos x = \dfrac{\sqrt2}{2}\\ \Leftrightarrow \sin x\cos\dfrac{\pi}{6} - \cos x\sin\dfrac{\pi}{6} = \sin\dfrac{\pi}{4}\\ \Leftrightarrow \sin\left(x - \dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{4}\\ \Leftrightarrow \left[\begin{array}{l}x - \dfrac{\pi}{6} = \dfrac{\pi}{4} +k2\pi\\x - \dfrac{\pi}{6} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =\dfrac{5\pi}{12} +k2\pi\\x = \dfrac{11\pi}{12} + k2\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
