Đáp án:
$\begin{array}{l}
a)\sqrt 3 .\sin x + \cos x = 2\cos \left( {x + \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin x + \dfrac{1}{2}\cos x = \cos \left( {x + \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \sin \dfrac{\pi }{3}.\sin x + \cos \dfrac{\pi }{3}.\cos x = \cos \left( {x + \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \cos \left( {x - \dfrac{\pi }{3}} \right) = \cos \left( {x + \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{3} = x + \dfrac{\pi }{3} + k2\pi \\
x - \dfrac{\pi }{3} = - x - \dfrac{\pi }{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow x = k\pi \\
Vậy\,x = k\pi \\
b)\sin x - \cos x = \sqrt 2 \cos \dfrac{\pi }{3}\\
\Leftrightarrow \sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = \sqrt 2 \cos \dfrac{\pi }{3}\\
\Leftrightarrow \cos \left( {\dfrac{\pi }{2} - x + \dfrac{\pi }{4}} \right) = \cos \dfrac{\pi }{3}\\
\Leftrightarrow \cos \left( {\dfrac{{3\pi }}{4} - x} \right) = \cos \dfrac{\pi }{3}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{3\pi }}{4} - x = \dfrac{\pi }{3} + k2\pi \\
\dfrac{{3\pi }}{4} - x = \dfrac{{ - \pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{12}} - k2\pi \\
x = \dfrac{{13\pi }}{{12}} - k2\pi
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{12}} - k2\pi \\
x = \dfrac{{13\pi }}{{12}} - k2\pi
\end{array} \right.
\end{array}$
