Đáp án:
a) $\left[\begin{array}{l}x = \pm \dfrac{\pi}{3} + k2\pi\\x = \dfrac{\pi}{4} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
b) $\left[\begin{array}{l}x = \dfrac{\pi}{6} +k2\pi\\x = \dfrac{5\pi}{6} +k2\pi\\x = \dfrac{\pi}{4}+ k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}a) \quad (2\cos x - 1)(2\sin x - \cos x) = \sin2x - \sin x\\ \Leftrightarrow (2\cos x - 1)(2\sin x - \cos x) = 2\sin x\cos x - \sin x\\ \Leftrightarrow (2\cos x - 1)(2\sin x - \cos x) = \sin x(2\cos x -1)\\ \Leftrightarrow (2\cos x -1)(2\sin x -\cos x - \sin x) =0\\ \Leftrightarrow (2\cos x -1)(\sin x - \cos x)=0\\ \Leftrightarrow (2\cos x -1).\sin\left(x - \dfrac{\pi}{4}\right) = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos x = \dfrac12\\\sin\left(x - \dfrac{\pi}{4}\right) = 0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \pm \dfrac{\pi}{3} + k2\pi\\x - \dfrac{\pi}{4} = k\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \pm \dfrac{\pi}{3} + k2\pi\\x = \dfrac{\pi}{4} + k\pi\end{array}\right.\quad (k\in\Bbb Z)\\ b)\quad (2\sin x - 1)(2\cos x - \sin x) = \sin2x - \cos x\\ \Leftrightarrow (2\sin x - 1)(2\cos x - \sin x) = 2\sin x\cos x - \cos x\\ \Leftrightarrow (2\sin x - 1)(2\cos x - \sin x) = \cos x(2\sin x -1)\\ \Leftrightarrow (2\sin x -1)(2\cos x - \sin x - \cos x) =0\\ \Leftrightarrow (2\sin x - 1)(\cos x - \sin x)=0\\ \Leftrightarrow (2\sin x - 1)\cos\left(x + \dfrac{\pi}{4}\right) =0\\ \Leftrightarrow \left[\begin{array}{l}\sin x = \dfrac12\\\cos\left(x + \dfrac{\pi}{4}\right) =0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} +k2\pi\\x = \dfrac{5\pi}{6} +k2\pi\\x + \dfrac{\pi}{4}=\dfrac{\pi}{2} + k\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} +k2\pi\\x = \dfrac{5\pi}{6} +k2\pi\\x = \dfrac{\pi}{4}+ k\pi\end{array}\right.\quad (k\in\Bbb Z) \end{array}$
