Đáp án:
$\begin{array}{l}
a){\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)^2} + \sqrt 3 \cos x = 2\\
\Leftrightarrow {\sin ^2}\dfrac{x}{2} + {\cos ^2}\dfrac{x}{2} + 2.\sin \dfrac{x}{2}.\cos \dfrac{x}{2} + \sqrt 3 \cos x = 2\\
\Leftrightarrow 1 + \sin x + \sqrt 3 \cos x = 2\\
\Leftrightarrow \sin x + \sqrt 3 \cos x = 1\\
\Leftrightarrow \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x = \dfrac{1}{2}\\
\Leftrightarrow \cos \dfrac{\pi }{3}.\sin x + \sin \dfrac{\pi }{3}.\cos x = \dfrac{1}{2}\\
\Leftrightarrow \sin \left( {x + \dfrac{\pi }{3}} \right) = \dfrac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{3} = \dfrac{\pi }{6} + k2\pi \\
x + \dfrac{\pi }{3} = \pi - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - \pi }}{6} + k2\pi \\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
Vậy\,x = \dfrac{{ - \pi }}{6} + k2\pi ;x = \dfrac{\pi }{2} + k2\pi \\
f)2\left( {\sqrt 3 \sin x - \cos x} \right) = 2\sin 2x + 2\sqrt 3 \cos 2x\\
\Leftrightarrow \sqrt 3 \sin x - \cos x = \sin 2x + \sqrt 3 \cos 2x\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x = \dfrac{1}{2}\sin 2x + \dfrac{{\sqrt 3 }}{2}\cos 2x\\
\Leftrightarrow \cos \left( {\dfrac{{ - \pi }}{6}} \right).\sin x + \sin \left( { - \dfrac{\pi }{6}} \right).\cos x\\
= \cos \left( {\dfrac{\pi }{3}} \right).\sin 2x + \sin \dfrac{\pi }{3}.\cos 2x\\
\Leftrightarrow \sin \left( {x - \dfrac{\pi }{6}} \right) = \sin \left( {2x + \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{6} = 2x + \dfrac{\pi }{3} + k2\pi \\
x - \dfrac{\pi }{6} = \pi - 2x - \dfrac{\pi }{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - \pi }}{2} - k2\pi \\
x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - \pi }}{2} - k2\pi ;x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array}$
