1 câu trả lời
Đáp án:
\(\left[ \begin{array}{l} x = \frac{\alpha }{2} - \frac{\pi }{6} + k\pi \\ x = - \frac{\alpha }{2} - \frac{\pi }{6} + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải: \(\begin{array}{l} 2{\cos ^2}x - \sqrt 3 \sin 2x = \sqrt 2 \\ \Leftrightarrow \cos 2x + 2 - \sqrt 3 \sin 2x = \sqrt 2 \\ \Leftrightarrow \cos 2x - \sqrt 3 \sin 2x = \sqrt 2 - 2\\ \Leftrightarrow \frac{1}{2}\cos 2x - \frac{{\sqrt 3 }}{2}\sin 2x = \frac{{\sqrt 2 - 2}}{2}\\ \Leftrightarrow \cos \left( {2x + \frac{\pi }{3}} \right) = \frac{{\sqrt 2 - 2}}{2} = \cos \alpha \\ \Leftrightarrow \left[ \begin{array}{l} 2x + \frac{\pi }{3} = \alpha + k2\pi \\ 2x + \frac{\pi }{3} = - \alpha + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 2x = \alpha - \frac{\pi }{3} + k2\pi \\ 2x = - \alpha - \frac{\pi }{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{\alpha }{2} - \frac{\pi }{6} + k\pi \\ x = - \frac{\alpha }{2} - \frac{\pi }{6} + k\pi \end{array} \right.\,\,\left( {k \in Z} \right). \end{array}\)
