Đáp án:
\(\begin{array}{l}
1)\,\,\,x = \frac{\pi }{4} + k\pi \,\,\,\left( {k \in Z} \right).\\
2)\,\,x = \frac{\pi }{{21}} + \frac{{k\pi }}{7}\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\,\,\sin 2x - 2{\cos ^2}2x - 1 = 0\\
\Leftrightarrow \sin 2x - 2\left( {1 - {{\sin }^2}2x} \right) - 1 = 0\\
\Leftrightarrow 2{\sin ^2}2x + \sin 2x - 3 = 0\\
\Leftrightarrow \left( {2\sin 2x + 3} \right)\left( {\sin 2x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 1\\
\sin 2x = - \frac{3}{2}\,\,\,\left( {ktm} \right)
\end{array} \right. \Leftrightarrow 2x = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{\pi }{4} + k\pi \,\,\,\left( {k \in Z} \right).\\
2)\,\,\sqrt 3 \cos 7x - 2\sin 5x.\cos 2x - \sin 3x = 0\\
\Leftrightarrow \,\sqrt 3 \cos 7x - \left( {\sin 3x + \sin 7x} \right) - \sin 3x = 0\\
\Leftrightarrow \sqrt 3 \cos 7x - \sin 3x - \sin 7x - \sin 3x = 0\\
\Leftrightarrow \sqrt 3 \cos 7x - \sin 7x = 0\\
\Leftrightarrow \frac{{\sqrt 3 }}{2}\cos 7x - \frac{1}{2}\sin 7x = 0\\
\Leftrightarrow \cos \left( {7x + \frac{\pi }{6}} \right) = 0\\
\Leftrightarrow 7x + \frac{\pi }{6} = \frac{\pi }{2} + k\pi \\
\Leftrightarrow x = \frac{\pi }{{21}} + \frac{{k\pi }}{7}\,\,\,\left( {k \in Z} \right).
\end{array}\)
