The number of solutions of \(\tan x+\sec x=2 \cos x\) in \([0,2 \pi]\) are

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