Đáp án:
\[x = \dfrac{\pi }{{12}} + k\pi \,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin 2x + \sqrt 3 \cos 2x = 2\\
\Leftrightarrow \dfrac{1}{2}\sin 2x + \dfrac{{\sqrt 3 }}{2}\cos 2x = 1\\
\Leftrightarrow \sin 2x.\cos \dfrac{\pi }{3} + \cos 2x.\sin \dfrac{\pi }{3} = 1\\
\Leftrightarrow \sin \left( {2x + \dfrac{\pi }{3}} \right) = 1\\
\Leftrightarrow 2x + \dfrac{\pi }{3} = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow 2x = \dfrac{\pi }{2} - \dfrac{\pi }{3} + k2\pi \\
\Leftrightarrow 2x = \dfrac{\pi }{6} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{{12}} + k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
