Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
\sin 2x - 2\sqrt 2 \sin x - \cos x + \sqrt 2 = 0\\
\Leftrightarrow 2\sin x.\cos x - 2\sqrt 2 \sin x - \cos x + \sqrt 2 = 0\\
\Leftrightarrow \left( {2\sin x.\cos x - 2\sqrt 2 \sin x} \right) - \left( {\cos x - \sqrt 2 } \right) = 0\\
\Leftrightarrow 2\sin x.\left( {\cos x - \sqrt 2 } \right) - \left( {\cos x - \sqrt 2 } \right) = 0\\
\Leftrightarrow \left( {\cos x - \sqrt 2 } \right)\left( {2\sin x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = \sqrt 2 \\
\sin x = \dfrac{1}{2}
\end{array} \right.\\
- 1 \le \cos x \le 1 \Rightarrow \cos x < \sqrt 2 \\
\Rightarrow \sin x = \dfrac{1}{2} \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\\
*)\\
{\sin ^2}x - \sqrt 3 \sin x.\cos x + 2{\cos ^2}x = 1\\
\Leftrightarrow {\sin ^2}x - \sqrt 3 \sin x.\cos x + 2{\cos ^2}x = {\sin ^2}x + {\cos ^2}x\\
\Leftrightarrow - \sqrt 3 \sin x.\cos x + {\cos ^2}x = 0\\
\Leftrightarrow \cos x.\left( {\cos x - \sqrt 3 \sin x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos x - \sqrt 3 \sin x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\dfrac{1}{2}\cos x - \dfrac{{\sqrt 3 }}{2}\sin x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos \left( {x + \dfrac{\pi }{3}} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x + \dfrac{\pi }{3} = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{6} + k\pi
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
