Đáp án:
$x = k\pi\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$(1 +\tan^2x)(1+\sin x) =\dfrac{\cos2x}{1 - \sin2x}\qquad (*)$
$ĐK:\,\begin{cases}\cos x \ne 0\\\sin2x \ne 1\end{cases}\Leftrightarrow x \ne \dfrac{\pi}{2} + n\pi$
$(*)\Leftrightarrow \dfrac{1+\sin x}{\cos^2x}=\dfrac{\cos^2x -\sin^2x}{\cos^2x -2\sin x\cos x + \sin^2x}$
$\Leftrightarrow \dfrac{1 +\sin x}{1 -\sin^2x}=\dfrac{(\cos x -\sin x)(\cos x +\sin x)}{(\cos x -\sin x)^2}$
$\Leftrightarrow \dfrac{1 +\sin x}{(1 -\sin x)(1+\sin x)}=\dfrac{\cos x +\sin x}{\cos x -\sin x}$
$\Leftrightarrow \cos x -\sin x = (\cos x +\sin x)(1-\sin x)$
$\Leftrightarrow \cos x -\sin x = \cos x -\sin x\cos x + \sin x -\sin ^2x$
$\Leftrightarrow \sin^2x - 2\sin x +\sin x\cos x = 0$
$\Leftrightarrow \sin x(\sin x +\cos x - 2) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x = 0\\\sin x +\cos x = 2\quad \text{(vô nghiệm)}\end{array}\right.$
$\Leftrightarrow x = k\pi\quad (k\in\Bbb Z)$
