Find out the general solution of $\tan x + \cot 2x = 2$. I want to solve for $x$ in

$$\tan x + \cot 2x = 2.$$

I tried to write in terms of $\sin x $ and   $\cos x$ but couldn't get the answer. Any help is appreciated.
$$\tan x+\cot2x=\dfrac{\sin x}{\cos x}+\dfrac{\cos2x}{2\sin x\cos x}=\dfrac{2\sin^2x+\cos2x}{2\sin x\cos x}=\csc2x$$  as $\cos2x=1-2\sin^2x$

$$\implies\sin2x=\dfrac12$$

Can you take it from here?
$\tan x+\cot 2x=2\implies(\sin x/\cos x)+(\cos2x/\sin2x)=0$.
Taking LCM,
$(\sin2x\sin x+\cos2x\cos x)/\sin2x\cos x=2$
{Using identity: $\cos(a-b)=\cos a\cos b+\sin a\sin b$}
$(\cos x/\sin 2x\cos x)=2\implies \sin2x=1/2=\sin(\pi/6) $
$2x=n\pi+(-1)^n (π/6)\implies x=nπ/2 +(-1)^n (π/12)$