Sinx + căn3 cosx = 2 sin2x

Đáp án: $\left[ \begin{array}{l}
x = \dfrac{\pi }{3} - k2\pi \\
x = \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)$

 Giải thích các bước giải:

$\begin{array}{l}
\sin x + \sqrt 3 \cos x = 2\sin 2x\\
 \Rightarrow \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x = \sin 2x\\
 \Rightarrow \cos \dfrac{\pi }{3}.\sin x + \sin \dfrac{\pi }{3}.\cos x = \sin 2x\\
 \Rightarrow \sin \left( {x + \dfrac{\pi }{3}} \right) = \sin 2x\\
 \Rightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{3} = 2x + k2\pi \\
x + \dfrac{\pi }{3} = \pi  - 2x + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} - k2\pi \\
x = \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$