Giải phương trình: \(2{\sin ^2}x - \sin x\cos x - {\cos ^2}x = 2\).

\(2{\sin ^2}x - \sin x\cos x - {\cos ^2}x = 2 \Leftrightarrow 1 - \cos 2x - \frac{1}{2}\sin 2x - \frac{{1 + \cos 2x}}{2} = 2\)

\( \Leftrightarrow 2 - 2\cos 2x - \sin 2x - 1 - \cos 2x = 4 \Leftrightarrow - 3\cos 2x - \sin 2x = 3\)

\( \Leftrightarrow \frac{{ - 3}}{{\sqrt {10} }}\cos 2x - \frac{1}{{\sqrt {10} }}\sin 2x = \frac{3}{{\sqrt {10} }}\)

Đặt \(\cos a = - \frac{3}{{\sqrt {10} }},\sin a = \frac{1}{{\sqrt {10} }}\)

\( \Leftrightarrow \cos \left( {a + 2x} \right) = \frac{3}{{\sqrt {10} }} \Leftrightarrow a + 2x = \arccos \frac{3}{{\sqrt {10} }} + k2\pi \) \(\left( {k \in \mathbb{Z}} \right)\)

Hoặc a + 2x = –arc\(\cos \frac{3}{{\sqrt {10} }} + k2\pi \) \(\left( {k \in \mathbb{Z}} \right)\)

\( \Rightarrow x = \arccos \frac{3}{{\sqrt {10} }} + k\pi - \frac{a}{2}\) hoặc 2x = –arc\(\cos \frac{3}{{2\sqrt {10} }} + k\pi - \frac{a}{2}\) \(\left( {k \in \mathbb{Z}} \right)\).