Giải thích các bước giải:
$\begin{array}{l}
1)\sin 2x - \sqrt 3 \cos 2x = 1\\
\Leftrightarrow \dfrac{1}{2}\sin 2x - \dfrac{{\sqrt 3 }}{2}\cos 2x = \dfrac{1}{2}\\
\Leftrightarrow \sin \left( {2x - \dfrac{\pi }{6}} \right) = \dfrac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{6} = \dfrac{\pi }{6} + k2\pi \\
2x - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
2){\cos ^2}2x + \dfrac{{\sqrt 3 }}{2}\sin 4x = 1\\
\Leftrightarrow \dfrac{{\cos 4x + 1}}{2} + \dfrac{{\sqrt 3 }}{2}\sin 4x = 1\\
\Leftrightarrow \dfrac{1}{2}\cos 4x + \dfrac{{\sqrt 3 }}{2}\sin 4x = \dfrac{1}{2}\\
\Leftrightarrow \cos \left( {4x - \dfrac{\pi }{3}} \right) = \dfrac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
4x - \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\
4x - \dfrac{\pi }{3} = - \dfrac{\pi }{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\dfrac{\pi }{2}\\
x = k\dfrac{\pi }{2}
\end{array} \right.\left( {k \in Z} \right)\\
3)1 + \sin x - \cos x - \sin 2x + 2\cos 2x = 0\\
\Leftrightarrow {\sin ^2}x + {\cos ^2}x + \sin x - \cos x - 2\sin x\cos x + 2\left( {{{\cos }^2}x - {{\sin }^2}x} \right) = 0\\
\Leftrightarrow \left( {{{\sin }^2}x + {{\cos }^2}x - 2\sin x\cos x} \right) + \left( {\sin x - \cos x} \right) + 2\left( {{{\cos }^2}x - {{\sin }^2}x} \right) = 0\\
\Leftrightarrow {\left( {\sin x - \cos x} \right)^2} + \left( {\sin x - \cos x} \right) + 2\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right) = 0\\
\Leftrightarrow \left( {\sin x - \cos x} \right)\left( {\sin x - \cos x + 1 - 2\left( {\cos x + \sin x} \right)} \right) = 0\\
\Leftrightarrow \left( {\sin x - \cos x} \right)\left( { - \sin x - 3\cos x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x - \cos x = 0\\
\sin x + 3\cos x = 1
\end{array} \right.\\
+ )TH1:\\
\sin x - \cos x = 0\\
\Leftrightarrow \sin x = \cos x\\
\Leftrightarrow \tan x = 1\\
\Leftrightarrow x = \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)\\
+ )TH2:\\
\sin x + 3\cos x = 1\\
\Leftrightarrow \dfrac{1}{{\sqrt {10} }}\sin x + \dfrac{3}{{\sqrt {10} }}\cos x = \dfrac{1}{{\sqrt {10} }}\\
\Leftrightarrow \sin \alpha \sin x + \cos \alpha \cos x = \sin \alpha \left( {\sin \alpha = \dfrac{1}{{\sqrt {10} }};\cos \alpha = \dfrac{3}{{\sqrt {10} }}} \right)\\
\Leftrightarrow \cos \left( {x - \alpha } \right) = \cos \left( {\dfrac{\pi }{2} - \alpha } \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x - \alpha = \dfrac{\pi }{2} - \alpha + k2\pi \\
x - \alpha = - \dfrac{\pi }{2} + \alpha + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = 2\alpha - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
Với ${\alpha :\sin \alpha = \dfrac{1}{{\sqrt {10} }};\cos \alpha = \dfrac{3}{{\sqrt {10} }}}$
