Đáp án:
$\begin{array}{l}
a)\dfrac{{\sqrt 3 }}{2}.\cos 4x + \dfrac{1}{2}.\sin 4x = \dfrac{{\sqrt 3 }}{2}\\
\Rightarrow \cos \dfrac{\pi }{6}.\cos 4x + \sin \dfrac{\pi }{6}.\sin 4x = \dfrac{{\sqrt 3 }}{2}\\
\Rightarrow \cos \left( {4x - \dfrac{\pi }{6}} \right) = \cos \dfrac{\pi }{6}\\
\Rightarrow \left[ \begin{array}{l}
4x - \dfrac{\pi }{6} = \dfrac{\pi }{6} + k2\pi \\
4x - \dfrac{\pi }{6} = - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{k\pi }}{2}
\end{array} \right.\\
b)\sin 2x - \cos 2x = - 1\\
\Rightarrow \sqrt 2 .\sin \left( {2x - \dfrac{\pi }{4}} \right) = - 1\\
\Rightarrow \sin \left( {2x - \dfrac{\pi }{4}} \right) = - \dfrac{1}{{\sqrt 2 }}\\
\Rightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{4} = \dfrac{{ - \pi }}{4} + k2\pi \\
2x - \dfrac{\pi }{4} = \dfrac{{5\pi }}{4} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{{3\pi }}{4} + k\pi
\end{array} \right.
\end{array}$
