Đặt \(\left\{{}\begin{matrix}\left|sinx\right|=a\ge0\\cosx=b\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}2b-a=1\\a^2+b^2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=\frac{a+1}{2}\\a^2+b^2=1\end{matrix}\right.\)
\(\Rightarrow a^2+\left(\frac{a+1}{2}\right)^2=1\)
\(\Leftrightarrow5a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=\frac{3}{5}\end{matrix}\right.\) \(\Rightarrow b=\frac{4}{5}\)
\(\Rightarrow cosx=\frac{4}{5}\Rightarrow x=\pm arccos\left(\frac{4}{5}\right)+k2\pi\)
