The number of solutions of the equation $ an x+\sec x=2 \cos x$ lying in the interval $[0,2 \pi]$ is

Question

The number of solutions of the equation $	an x+\sec x=2 \cos x$ lying in the interval $[0,2 \pi]$ is

Aicharya · Accepted Answer

Given $	an x+\sec x=2 \cos x$ $\frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x \Rightarrow \sin x+1=2 \cos ^{2} x$ $\sin +1=2\left(1-\sin ^{2} xight) \Rightarrow 2 \sin ^{2} x+\sin x-1=0$ $(2 \sin x-1)(\sin x+1)=0 \Rightarrow \sin x=\frac{1}{2}, \sin x=-1$ If $\sin x=-1$, then $x=\frac{3 \pi}{2}$ and $\cos \frac{3 \pi}{2}=0$. Hence given equation is not defined at $\sin x=-1$. $	herefore \sin x=\frac{1}{2} \Rightarrow x=\frac{\pi}{6}, \frac{5 \pi}{6}$

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

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