sin^2x=1/2 sin3x+sin( pi/4-x/2)=0

Đáp án:

Giải thích các bước giải: a) $\begin{array}{l} {\sin ^2}x = \frac{1}{2} \Leftrightarrow \frac{{1 - \cos 2x}}{2} = \frac{1}{2} \Leftrightarrow \cos 2x = 0\\ \Leftrightarrow 2x = \frac{\pi }{2} + k\pi \Leftrightarrow x = \frac{\pi }{4} + \frac{{k\pi }}{2} \end{array}$ b) $\begin{array}{l} \sin 3x + \sin \left( {\frac{\pi }{4} - \frac{x}{2}} \right) = 0 \Leftrightarrow \sin 3x = - \sin \left( {\frac{\pi }{4} - \frac{x}{2}} \right)\\ \Leftrightarrow \sin 3x = \sin \left( { - \frac{\pi }{4} + \frac{x}{2}} \right) \Leftrightarrow \left[ \begin{array}{l} 3x = - \frac{\pi }{4} + \frac{x}{2} + k2\pi \\ 3x = \pi + \frac{\pi }{4} - \frac{x}{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \frac{{5x}}{2} = - \frac{\pi }{4} + k2\pi \\ \frac{{7x}}{2} = \frac{{5\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \frac{\pi }{{10}} + \frac{{k4\pi }}{5}\\ x = \frac{{5\pi }}{{14}} + \frac{{k4\pi }}{7} \end{array} \right. \end{array}$