Đáp án:
$ \left\{ \begin{array}{l} x = k2\pi \\ x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3} \end{array} \right.$ $(k\in\mathbb Z)$
Lời giải:
$\begin{array}{l} (2\cos x - 1)(\sin x + \cos x)=1\\ \Leftrightarrow 2\sin x\cos x - \sin x - \cos x + 2{\cos ^2}x - 1 = 0\\ \Leftrightarrow \sin 2x - \sin x - \cos x + \cos 2x = 0\\ \Leftrightarrow \sin 2x + \cos 2x = \sin x + \cos x\\ \Leftrightarrow \sqrt 2 \sin \left( {2x + \dfrac{\pi }{4}} \right) = \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right)\\ \Leftrightarrow \sin \left( {2x + \dfrac{\pi }{4}} \right) = \sin \left( {x + \dfrac{\pi }{4}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} 2x + \dfrac{\pi }{4} = x + \dfrac{\pi }{4} + k2\pi \\ 2x + \dfrac{\pi }{4} = \pi - x - \dfrac{\pi }{4} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = k2\pi \\ x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3} \end{array} \right. \end{array}$
Vậy phương trình có nghiệm $ \left\{ \begin{array}{l} x = k2\pi \\ x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3} \end{array} \right.$ $(k\in\mathbb Z)$
