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Phương trình đã cho tương đương với:
$\cos2x+\sqrt3\sin2x=1$
$\Leftrightarrow \dfrac{1}{2}\cos2x+\dfrac{\sqrt3}{2}\sin2x=\dfrac{1}{2}$
$\Leftrightarrow \sin(2x+\dfrac{\pi}{6})=\sin\dfrac{\pi}{6}$
$\Leftrightarrow \left[\begin{array}{l}2x+\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\x+\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=k\pi\\x=\dfrac{\pi}{3}+k\pi\end{array}\right.,k\in\mathbb{Z}$
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