Giải thích các bước giải:
Áp dụng \(\cos 2x = 2{\cos ^2}x - 1 \Leftrightarrow {\cos ^2}x = \frac{{\cos 2x + 1}}{2}\) ta có:
\(\begin{array}{l}
{\sin ^2}\frac{x}{2} - {\cos ^2}\left( {\frac{x}{4}} \right) + \frac{3}{4} = 0\\
\Leftrightarrow {\sin ^2}\frac{x}{2} - \frac{{\cos \frac{x}{2} + 1}}{2} + \frac{3}{4} = 0\\
\Leftrightarrow 1 - {\cos ^2}\frac{x}{2} - \frac{1}{2}\cos \frac{x}{2} - \frac{1}{2} + \frac{3}{4} = 0\\
\Leftrightarrow - {\cos ^2}\frac{x}{2} - \frac{1}{2}\cos \frac{x}{2} + \frac{5}{4} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos \frac{x}{2} = \frac{{ - 1 + \sqrt {21} }}{4}\\
\cos \frac{x}{2} = \frac{{ - 1 - \sqrt {21} }}{4}
\end{array} \right. \Rightarrow \cos \frac{x}{2} = \frac{{ - 1 + \sqrt {21} }}{4}
\end{array}\)
