a) Sin2x =1/2 b) 2sinx+4cosx=2

Đáp án:

\(\begin{array}{l} a)\,\,\left[ \begin{array}{l} x = \frac{\pi }{{12}} + k\pi \\ x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\\ b)\,\,\,\left[ \begin{array}{l} x = \frac{\pi }{2} - 2\alpha + k2\pi \\ x = \frac{\pi }{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right) \end{array}\)

Giải thích các bước giải: \(\begin{array}{l} a)\,\,\sin 2x = \frac{1}{2}\\ \Leftrightarrow \left[ \begin{array}{l} 2x = \frac{\pi }{6} + k2\pi \\ 2x = \frac{{5\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{{12}} + k\pi \\ x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\\ b)\,\,\,2\sin x + 4\cos x = 2\\ \Leftrightarrow \sin x + 2\cos x = 1\\ \Leftrightarrow \frac{1}{{\sqrt 5 }}\sin x + \frac{2}{{\sqrt 5 }}\cos x = \frac{1}{{\sqrt 5 }}\\ \Leftrightarrow \sin \left( {x + \alpha } \right) = \sin \left( {\frac{\pi }{2} - \alpha } \right)\,\,\,\,\left( {voi\,\,\,\sin \alpha = \frac{2}{{\sqrt 5 }};\,\,\,\cos \alpha = \frac{1}{{\sqrt 5 }}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} x + \alpha = \frac{\pi }{2} - \alpha + k2\pi \\ x + \alpha = \frac{\pi }{2} + \alpha + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{2} - 2\alpha + k2\pi \\ x = \frac{\pi }{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right) \end{array}\)