1) sin2x-2cos^2 2x-1=0 2) căn3 cos7x-2sin5xcos2x-sin3x=0

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