Given trigonometric equation is,

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

$\Rightarrow \frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }=2\cos \theta$

$\Rightarrow 1+\sin \theta =2{\cos }^2\theta \left[\cos \theta \ne 0;\theta \ne \frac{\pi }{2},\frac{3\pi }{2} \text{in} \left[0,2\pi \right]\right]$

$\Rightarrow 1+\sin \theta =2\left(1-{\sin }^2\theta \right)$

$\Rightarrow 2{\sin }^2\theta +\sin \theta -1=0$

$\Rightarrow 2{\sin }^2\theta +2\sin \theta -\sin \theta -1=0$

$\Rightarrow \left(2\sin \theta -1\right)\left(\sin \theta +1\right)=0$

$\therefore$Either $\sin \theta =\frac{1}{2} \text{or} \sin \theta =-1$

$\therefore \theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}$

But $\theta$ can not be $\frac{3\pi }{2}$

$\therefore \theta =\frac{\pi }{6},\frac{5\pi }{6}$

Hence, number of solutions are $2$.