Question

√3-ctg^2(x)-1/2ctgx=0

Answer 1

$\frac{ctg^2x-1}{2ctgx}= (\frac{\cos^2x-\sin^2x}{\sin^2x}):\frac{2\cos x}{\sin x}=\frac{\cos2x}{\sin^2x}\cdot\frac{\sin x}{2\cos x}=\frac{\cos2x}{2\sin x\cos x}=\frac{\cos2x}{\sin2x}=ctg 2x.$

$ctg2x=\sqrt3;\\\\2x=arcctg(\sqrt3)+\pi k, k\in \mathbb Z;\\\\2x=\frac{\pi}{6}+\pi k, k\in \mathbb Z\\\\x=\frac{\pi}{12}+\frac{\pi k}{2} , k\in \mathbb Z$