The number of solutions of $\tan x+secx=2\cos x$ in [0,$2\pi$ ] is ____________

The given equation is $\tan x+secx=2\cos x$

This can be written as $\frac{\sin x}{\cos x}+\frac{1}{\cos x}=2\cos x$

Here the condition  is $\cos x\ne 0$

and the equation is 

   $\begin{array}{rcl}1+\sin x& =& 2{\cos }^{2}x\\ & =& 2\left(1-{\sin }^{2}x\right)\\ & =& 2\left(1-\sin x\right)\left(1+\sin x\right)\end{array}$

It implies that $1+\sin x=0\Rightarrow \sin x=-1\Rightarrow x=\frac{3\mathrm{\pi }}{2}$, but this is not solution for the given equaiton because $\cos \left(\frac{3\mathrm{\pi }}{2}\right)$ is zero. 

And 

$\begin{array}{rcl}1-\sin x& =& \frac{1}{2}\\ \sin x& =& \frac{1}{2}\\ x& =& \frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6}\end{array}$

Therefore, the number of solutions for the given equation is $\boxed{\mathbf{2}}$