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

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