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

The number of solutions of $\tan x+\sec x=2 \cos x$ in $[0,2 \pi]$ is
  1. 2
  2. 3
  3. 0
  4. 1

Solution

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

Asked in: MHT CET 2024 (11 May Shift 2)

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