Đáp án:

$\left[\begin{array}{l} x=k 2 \pi(k \in \mathbb{Z}) \\ x=\dfrac{\pi}{4}+k  \pi (k \in \mathbb{Z}) \end{array} \right..$

Giải thích các bước giải:

$\tan^2x=\dfrac{1-\cos x}{1-\sin x}\\ \text{ĐKXĐ: }\left\{\begin{array}{l} x \ne \dfrac{\pi}{2}+ k \pi(k \in \mathbb{Z} )\\ \sin x \ne 1\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ne \dfrac{\pi}{2}+ k \pi(k \in \mathbb{Z}) \\ x \ne \dfrac{\pi}{2}+ k2 \pi(k \in \mathbb{Z})\end{array} \right.\\ \Leftrightarrow x \ne \dfrac{\pi}{2}+ k \pi(k \in \mathbb{Z})\\ \tan^2x=\dfrac{1-\cos x}{1-\sin x}\\ \Leftrightarrow \tan^2x=\dfrac{(1-\cos x)(1+\sin x)}{(1-\sin x)(1+\sin x)}\\ \Leftrightarrow \tan^2x=\dfrac{\sin x-\cos x-\sin x \cos x+1}{\cos^2x}\\ \Leftrightarrow \sin^2x=\sin x-\cos x-\sin x \cos x+1\\ \Leftrightarrow \sin^2x-\sin x+\cos x+\sin x \cos x-1\\ \Leftrightarrow -\cos^2x-\sin x+\cos x+\sin x \cos x=0\\ \Leftrightarrow -\cos^2x+\cos x-\sin x+\sin x \cos x=0\\ \Leftrightarrow \cos x(1-\cos x+)-\sin x(1-\cos x)=0\\ \Leftrightarrow (1-\cos x)(\cos x-\sin x)=0\\ \Leftrightarrow \left[\begin{array}{l} \cos x=1 \\ \cos x=\sin x\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=k 2 \pi(k \in \mathbb{Z}) \\ \cos x=\cos \left(\dfrac{\pi}{2}-x\right)\end{array} \right. \\ \Leftrightarrow \left[\begin{array}{l} x=k 2 \pi(k \in \mathbb{Z}) \\ x=\dfrac{\pi}{2}-x+k 2 \pi (k \in \mathbb{Z}) \\ x=x-\dfrac{\pi}{2}+k 2 \pi (k \in \mathbb{Z})\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=k 2 \pi(k \in \mathbb{Z}) \\ 2x=\dfrac{\pi}{2}+k 2 \pi (k \in \mathbb{Z}) \\ 0=-\dfrac{\pi}{2}+k 2 \pi (k \in \mathbb{Z})(L)\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=k 2 \pi(k \in \mathbb{Z}) \\ x=\dfrac{\pi}{4}+k  \pi (k \in \mathbb{Z}) \end{array} \right..$