Đáp án:
\[\left[ \begin{array}{l}
x = \frac{1}{2}\arcsin \frac{3}{4} + k\pi \\
x = \frac{\pi }{2} - \frac{1}{2}\arcsin \frac{3}{4} + k\pi \\
x = \frac{\pi }{{12}} + m\pi \\
x = \frac{{5\pi }}{6} + m\pi
\end{array} \right.\,\,\left( {k,\,\,m \in Z} \right).\]
Giải thích các bước giải:
\[\begin{array}{l}
4\cos 4x + 10\sin 2x - 7 = 0\\
\Leftrightarrow 4\left( {1 - 2{{\sin }^2}2x} \right) + 10\sin 2x - 7 = 0\\
\Leftrightarrow 4 - 8{\sin ^2}2x + 10\sin 2x - 7 = 0\\
\Leftrightarrow 8{\sin ^2}2x - 10\sin 2x + 3 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = \frac{3}{4}\\
\sin 2x = \frac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = \arcsin \frac{3}{4} + k2\pi \\
2x = \pi - \arcsin \frac{3}{4} + k2\pi \\
2x = \frac{\pi }{6} + m2\pi \\
2x = \frac{{5\pi }}{6} + m2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{1}{2}\arcsin \frac{3}{4} + k\pi \\
x = \frac{\pi }{2} - \frac{1}{2}\arcsin \frac{3}{4} + k\pi \\
x = \frac{\pi }{{12}} + m\pi \\
x = \frac{{5\pi }}{6} + m\pi
\end{array} \right.\,\,\left( {k,\,\,m \in Z} \right).
\end{array}\]
