The number of solutions of $ an x+\sec x=2 \cos x$ in $[0,2 \pi]$ are

Question

The number of solutions of $	an x+\sec x=2 \cos x$ in $[0,2 \pi]$ are

Aicharya · Accepted Answer

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

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

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