Giải thích các bước giải:
$\begin{array}{l}
a)\sin 3x + \cos 3x = \sqrt 2 \\
\Rightarrow \sqrt 2 .\sin \left( {3x + \dfrac{\pi }{4}} \right) = \sqrt 2 \\
\Rightarrow \sin \left( {3x + \dfrac{\pi }{4}} \right) = 1\\
\Rightarrow 3x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \\
\Rightarrow 3x = \dfrac{\pi }{4} + k2\pi \\
\Rightarrow x = \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}\\
Vay\,x = \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}\\
b)2{\sin ^2}x + \sqrt 3 \sin 2x = 3\\
\Rightarrow 2{\sin ^2}x - 1 + \sqrt 3 \sin 2x = 2\\
\Rightarrow - \cos 2x + \sqrt 3 \sin 2x = 2\\
\Rightarrow \dfrac{{\sqrt 3 }}{2}\sin 2x - \dfrac{1}{2}\cos 2x = 1\\
\Rightarrow \sin \left( {2x - \dfrac{\pi }{3}} \right) = 1\\
\Rightarrow 2x - \dfrac{\pi }{3} = \dfrac{\pi }{2} + k2\pi \\
\Rightarrow x = \dfrac{{5\pi }}{{12}} + k\pi \\
Vay\,x = \dfrac{{5\pi }}{{12}} + k\pi \\
c)\\
\sqrt 3 \cos \left( {x + \dfrac{\pi }{2}} \right) + \sin \left( {x - \dfrac{\pi }{2}} \right) = 2\sin 2x\\
\Rightarrow \sqrt 3 \sin x + \cos x = 2\sin 2x\\
\Rightarrow \dfrac{{\sqrt 3 }}{2}\sin x + \dfrac{1}{2}\cos x = \sin 2x\\
\Rightarrow \sin \left( {x + \dfrac{\pi }{3}} \right) = \sin 2x\\
\Rightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{3} = 2x + k2\pi \\
x + \dfrac{\pi }{3} = \pi - 2x + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} - k2\pi \\
x = \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.
\end{array}$
