The number of solutions of the equation $\tan x + secx = 2\cos x$ lying in the interval [0, 2$\mathrm{\pi }$] is

$\tan x+\sec x=2\cos x$---------(i)
is meaningful when $\cos x\ne 0.\text{ i.e., }\sin x\ne 1\text{ or }-1$
From eq. (i), we have
$\begin{array}{l}\sin x+1=2{\cos }^{2}x=2\left(1-{\sin }^{2}x\right)\\ \Rightarrow 2{\sin }^{2}x+\sin x-1=0\\ \Rightarrow (1+\sin x)(2\sin x-1)=0\\ \Rightarrow \sin x=\frac{1}{2},[\because \sin x\ne -1]\\ \Rightarrow x=\frac{\pi }{6},\frac{5\pi }{6}\text{ in }[0,2\pi ].\end{array}$