Giúp em bài 1 ạ, cảm ơn trước

Đáp án:

$\begin{array}{l}
1) - 4{\cos ^2}x + 2\left( {\sqrt 3  - 1} \right)\sin x + 4 - \sqrt 3  = 0\\
 \Rightarrow  - 4\left( {1 - {{\sin }^2}x} \right) + 2\left( {\sqrt 3  - 1} \right)\sin x + 4 - \sqrt 3  = 0\\
 \Rightarrow 4{\sin ^2}x + 2\left( {\sqrt 3  - 1} \right)\sin x - \sqrt 3  = 0\\
 \Rightarrow \left[ \begin{array}{l}
\sin x = \frac{1}{2}\\
\sin x = \frac{{ - \sqrt 3 }}{2}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \\
x = \frac{{5\pi }}{6} + k2\pi \\
x = \frac{{ - \pi }}{3} + k2\pi \\
x = \frac{{4\pi }}{3} + k2\pi 
\end{array} \right.\\
2)\cos 2\left( {x + \frac{\pi }{4}} \right) + 4\cos \left( {\frac{\pi }{4} - x} \right) = \frac{5}{2}\\
 \Rightarrow 1 - 2{\sin ^2}\left( {x + \frac{\pi }{4}} \right) + 4\sin \left( {\frac{\pi }{4} - x} \right) = \frac{5}{2}\\
 \Rightarrow  - 2{\sin ^2}\left( {x + \frac{\pi }{4}} \right) + 4\sin \left( {\frac{\pi }{4} - x} \right) - \frac{3}{2} = 0\\
 \Rightarrow \sin \left( {\frac{\pi }{4} - x} \right) = \frac{1}{2}\\
 \Rightarrow \left[ \begin{array}{l}
\frac{\pi }{4} - x = \frac{\pi }{6} + k2\pi \\
\frac{\pi }{4} - x = \frac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{12}} - k2\pi \\
x = \frac{{ - 7\pi }}{{12}} - k2\pi 
\end{array} \right.
\end{array}$