Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^4}\frac{x}{2} + {\cos ^4}\frac{x}{2} = 1 - 2\sin x\\
\Leftrightarrow {\left( {{{\sin }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2}} \right)^2} - 2{\sin ^2}\frac{x}{2}.{\cos ^2}\frac{x}{2} = 1 - 2\sin x\\
\Leftrightarrow 1 - \frac{1}{2}.{\left( {2.\sin \frac{x}{2}.\cos \frac{x}{2}} \right)^2} = 1 - 2\sin x\\
\Leftrightarrow \frac{1}{2}.si{n^2}x = 2\sin x\\
\Leftrightarrow {\sin ^2}x - 4\sin x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = 4
\end{array} \right. \Rightarrow \sin x = 0 \Leftrightarrow x = k\pi
\end{array}\)
