The given equation is defined for $x \neq \frac{\pi}{2}, \frac{3 \pi}{2}$.
Now, $\tan x+\sec x=2 \cos x$
$$
\begin{aligned}
& \Rightarrow \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x \\
& \Rightarrow(\sin x+1)=2 \cos ^2 x \\
& \Rightarrow(\sin x+1)=2\left(1-\sin ^2 x\right) \\
& \Rightarrow(\sin x+1)=2(1-\sin x)(1+\sin x) \\
& \Rightarrow(1+\sin x)[2(1-\sin x)-1]=0 \\
& \Rightarrow 2(1-\sin x)-1=0 \\
& \quad \ldots\left[\begin{array}{r}
\sin x \neq-1 \text { otherwise } \cos x=0 \text { and } \\
\tan x, \sec x \text { will be undefined }
\end{array}\right] \\
& \Rightarrow \sin x=\frac{1}{2} \quad \\
& \Rightarrow x=\frac{\pi}{6}, \frac{5 \pi}{6} \text { in }(0,2 \pi)
\end{aligned}
$$