MHT CET · Maths · Functions
The number of solutions of the equation \(\tan x+\sec x=2 \cos x\) lying in the interval \([0,2 \pi]\) is
- A 0
- B 2
- C 3
- D 1
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
Given
\(\tan x+\sec x=2 \cos x\)
\(\frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x \Rightarrow \sin x+1=2 \cos ^{2} x\)
\(\sin +1=2\left(1-\sin ^{2} x\right) \Rightarrow 2 \sin ^{2} x+\sin x-1=0\)
\((2 \sin x-1)(\sin x+1)=0 \Rightarrow \sin x=\frac{1}{2}, \sin x=-1\)
If \(\sin x=-1\), then \(x=\frac{3 \pi}{2}\) and \(\cos \frac{3 \pi}{2}=0\).
Hence given equation is not defined at \(\sin x=-1\).
\(\therefore \sin x=\frac{1}{2} \Rightarrow x=\frac{\pi}{6}, \frac{5 \pi}{6}\)
\(\tan x+\sec x=2 \cos x\)
\(\frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x \Rightarrow \sin x+1=2 \cos ^{2} x\)
\(\sin +1=2\left(1-\sin ^{2} x\right) \Rightarrow 2 \sin ^{2} x+\sin x-1=0\)
\((2 \sin x-1)(\sin x+1)=0 \Rightarrow \sin x=\frac{1}{2}, \sin x=-1\)
If \(\sin x=-1\), then \(x=\frac{3 \pi}{2}\) and \(\cos \frac{3 \pi}{2}=0\).
Hence given equation is not defined at \(\sin x=-1\).
\(\therefore \sin x=\frac{1}{2} \Rightarrow x=\frac{\pi}{6}, \frac{5 \pi}{6}\)
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