ĐKXĐ: \(sinx\ne0\)
\(\Leftrightarrow cot^2x=\left|\frac{1-\left|sinx\right|}{1-\left|cosx\right|}\right|\Leftrightarrow cot^2x=\frac{1-\left|sinx\right|}{1-\left|cosx\right|}\)
Đặt \(\left\{{}\begin{matrix}\left|sinx\right|=a\Rightarrow0< a\le1\\\left|cosx\right|=b\Rightarrow0\le b< 1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\frac{b^2}{a^2}=\frac{1-a}{1-b}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\frac{b^2}{1-b^2}=\frac{1-a}{1-b}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\frac{b^2}{\left(1-b\right)\left(1+b\right)}=\frac{\left(1-a\right)\left(1+b\right)}{\left(1-b\right)\left(1+b\right)}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\1-a^2=\left(1-a\right)\left(1-b\right)\end{matrix}\right.\)
\(\Rightarrow\left(1-a\right)\left(1+a\right)=\left(1-a\right)\left(1-b\right)\)
\(\Rightarrow\left[{}\begin{matrix}a=1\\1+a=1-b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left|sinx\right|=1\Leftrightarrow sin^2x=1\Leftrightarrow cosx=0\Leftrightarrow x=...\)
