Giải thích các bước giải:
$\dfrac{\sin(x)}{1+\cos(x)}+\dfrac{1}{1-\cos(x)}+\cot(x)=2$
$\to\dfrac{\sin(x).(1-\cos(x))+1+\cos(x)}{(1+\cos(x))(1-\cos(x))}+\dfrac{\cos(x)}{\sin(x)}=2$
$\to\dfrac{\sin(x)-\sin(x)\cos(x)+1+\cos(x)}{1-\cos^2(x)}+\dfrac{\cos(x)}{\sin(x)}=2$
$\to\dfrac{\sin(x)-\sin(x)\cos(x)+1+\cos(x)}{\sin^2(x)}+\dfrac{\cos(x)}{\sin(x)}=2$
$\to\sin(x)-\sin(x)\cos(x)+1+\cos(x)+\cos(x).\sin(x)=2\sin^2(x)$
$\to\sin(x)+1+\cos(x)=2\sin^2(x)$
$\to\sin(x)+\cos(x)=2\sin^2(x)-1$
$\to\sin(x)+\cos(x)=\sin^2(x)-\cos^2(x)$
$\to\sin(x)+\cos(x)=(\sin(x)+\cos(x))(\sin(x)-\cos(x))$
$\to(\sin(x)+\cos(x))(\sin(x)-\cos(x)-1)=0$
+) $\sin(x)+\cos(x)=0\to x=\dfrac{3\pi}{4}+k\pi$
+) $\sin(x)-\cos(x)-1=0$
$\to \cos \left(\dfrac{\pi }{4}+x\right)=-\dfrac{\sqrt{2}}{2}$
$\to x=2\pi k+\dfrac{\pi }{2},\:x=2\pi k+\pi $
