The number of solutions of the equation $\tan x+\sec x=2\cos x,x\in [0,2\pi ]$ is

$\tan x+\sec x=2\cos x$

$\begin{array}{l}\Rightarrow 1+\sin x=2{\cos }^{2}x=2\left(1-{\sin }^{2}x\right)\\ \Rightarrow (1+\sin x)(1-2+2\sin x)=0\\ \Rightarrow \sin x=-1\text{ or }\sin x=1/2\end{array}$

But $\sin x=-1\Rightarrow \cos x=0$ This is not possible.

Thus $\sin x=1/2\Rightarrow x=\pi /6,5\pi /6,\in [0,2\pi ]$