Đặt \(\left\{{}\begin{matrix}\left|sin\frac{x}{2}\right|=a\\\left|cos\frac{x}{2}\right|=b\end{matrix}\right.\) với \(a;b\in\left[0;1\right]\)
\(\left\{{}\begin{matrix}a^2+b^2=1\\a^4-b^4=a-b\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\left(a^2-b^2\right)\left(a^2+b^2\right)=a-b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\left(a-b\right)\left(a+b\right)=a-b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\left[{}\begin{matrix}a=b\\a+b=1\end{matrix}\right.\end{matrix}\right.\)
TH1: \(a=b\Leftrightarrow\left|sin\frac{x}{2}\right|=\left|cos\frac{x}{2}\right|\)
\(\Leftrightarrow cos^2\frac{x}{2}=sin^2\frac{x}{2}\Leftrightarrow cos^2\frac{x}{2}-sin^2\frac{x}{2}=0\)
\(\Leftrightarrow cosx=0\Leftrightarrow x=...\)
TH2: \(\left\{{}\begin{matrix}a^2+b^2=1\\a+b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\left(a+b\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow ab=0\Leftrightarrow\left|sin\frac{x}{2}.cos\frac{x}{2}\right|=0\)
\(\Leftrightarrow sinx=0\Leftrightarrow x=...\)
