Number of solutions of equation $\tan^{-1}(2\sin x)=\cot^{-1}(\cos x)\;$ in $[0,10\pi]$.

Question

Number of solutions of equation $\tan^{-1}(2\sin x)=\cot^{-1}(\cos x)\;$ in $[0,10\pi]$.

My Approach:

Using the formula $\tan^{-1}(x)\;+\;\cot^{-1}(x)=\dfrac{\pi}{2}$ our equation get converted to

$\tan^{-1}(2\sin x)+\tan^{-1}(\cos x)\;=\dfrac{\pi}{2}$

S0 $2\sin x \cos x=1$

$\implies \sin (2x)=1$

$\implies 2x=(4n+1)\dfrac{\pi}{2}\; \implies \; x=n\pi\;+\;\dfrac{\pi}{4}$

But answer given is $x=2n\pi\;+\;\dfrac{\pi}{4}$.

Where Am i going wrong?