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) $\begin{array}{l}
\sin 3x + \sin \left( {\frac{\pi }{4} – \frac{x}{2}} \right) = 0 \Leftrightarrow \sin 3x = – \sin \left( {\frac{\pi }{4} – \frac{x}{2}} \right)\\
\Leftrightarrow \sin 3x = \sin \left( { – \frac{\pi }{4} + \frac{x}{2}} \right) \Leftrightarrow \left[ \begin{array}{l}
3x = – \frac{\pi }{4} + \frac{x}{2} + k2\pi \\
3x = \pi + \frac{\pi }{4} – \frac{x}{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\frac{{5x}}{2} = – \frac{\pi }{4} + k2\pi \\
\frac{{7x}}{2} = \frac{{5\pi }}{4} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – \frac{\pi }{{10}} + \frac{{k4\pi }}{5}\\
x = \frac{{5\pi }}{{14}} + \frac{{k4\pi }}{7}
\end{array} \right.
\end{array}$