Điều kiện: $sin(\frac{3\pi}{2}+x)\neq 0\Rightarrow x\neq -\frac{3\pi}{2}+k\pi(k\in Z)$$cos(x)\neq 0\Rightarrow x\neq \frac{\pi}{2}+k\pi(k\in Z)$
Ta có: $cot(\frac{3\pi}{2}+x)=-tan(\pi+x)=-tan(x)$
Ptr ban đầu: $-tan(x)-tan^2(x)=\frac{cos(2x)-1}{cos^2(x)}$
$\Leftrightarrow \frac{cos(2x)-1}{cos^2(x)}+\frac{sin^2(x)}{cos^2(x)}+\frac{sin(x)}{cos(x)}=0$
$\Leftrightarrow cos(2x)-1+sin^2(x)+sin(x)cos(x)=0$
$\Leftrightarrow 1-2sin^2(x)-1+sin^2(x)+sin(x)cos(x)=0$
$\Leftrightarrow sin(x)cos(x)-sin^2(x)=0$
$\Leftrightarrow sin(x)[cos(x)-sin(x)]=0$
$\Leftrightarrow sin(x)=0$ hay $sin(x)=cos(x)$
Với $sin(x)=0\Rightarrow x=k\pi(k\in Z)$
Với $sin(x)-cos(x)=0\Rightarrow sin(x-\frac{\pi}{4})=0\Rightarrow x=\frac{\pi}{4}+k\pi(k\in Z)$