Giải phương trình sin2x+√3 cos2x =√3

Đáp án: $\left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{6} + k\pi 
\end{array} \right.\left( {k \in Z} \right)$

Giải thích các bước giải:

$\begin{array}{l}
\sin 2x + \sqrt 3 \cos 2x = \sqrt 3 \\
 \Leftrightarrow \dfrac{1}{2}\sin 2x + \dfrac{{\sqrt 3 }}{2}\cos 2x = \dfrac{{\sqrt 3 }}{2}\\
 \Leftrightarrow \sin 2x.\cos \dfrac{\pi }{3} + \sin \dfrac{\pi }{3}.\cos 2x = \dfrac{{\sqrt 3 }}{2}\\
 \Leftrightarrow \sin \left( {2x + \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\
2x + \dfrac{\pi }{3} = \pi  - \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = k2\pi \\
2x = \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{6} + k\pi 
\end{array} \right.\left( {k \in Z} \right)\\
Vậy\,\left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{6} + k\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}$