giải phương trình : 2cot2x-cotx=tan(x/2)

Giải thích các bước giải:

$2\cot 2x-\cot x=\tan \dfrac{x}{2}$ 

$\to \cot 2x-\cot x=\tan \dfrac{x}{2}-\cot 2x$ 

$\to \dfrac{\cos 2x}{\sin 2x}-\dfrac{\cos x}{\sin x}=\dfrac{\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}}-\dfrac{\cos 2x}{\sin 2x}$ 

$\to \dfrac{\cos 2x.\sin x-\sin 2x.\cos x}{\sin 2x.\sin x}=\dfrac{\sin\dfrac{x}{2}.\sin 2x-\cos\dfrac{x}{2}.\cos 2x}{\cos\dfrac{x}{2}.\sin 2x}$ 

$\to \dfrac{-\sin x}{\sin 2x.\sin x}=-\dfrac{\cos\dfrac{5x}{2}}{\cos\dfrac{x}{2}.\sin 2x}$ 

$\to\dfrac{\cos\dfrac{5x}{2}}{\cos\dfrac{x}{2}}=1$ 

$\to \cos\dfrac{5x}{2}=\cos\dfrac{x}{2}$

$\to x=\dfrac{4\pi n}{3},\:x=\dfrac{2\pi }{3}+\dfrac{4\pi n}{3},\:x=2\pi n,\:x=\pi +2\pi n$