MHT CET · Maths · Trigonometric Equations
The principal solutions of the equation \(\sec x+\tan x=2 \cos x\) are
- A \(\frac{\pi}{6}, \frac{5 \pi}{6}\)
- B \(\frac{\pi}{6}, \frac{\pi}{20}\)
- C \(\frac{\pi}{6}, \frac{2 \pi}{3}\)
- D \(\frac{\pi}{6}, \frac{\pi}{12}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{6}, \frac{5 \pi}{6}\)
Step-by-step Solution
Detailed explanation
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 \)
\( \Rightarrow(\sin x+1)=2 \cos ^2 x \)
\( \Rightarrow(\sin x+1)=2\left(1-\sin ^2 x\right) \)
\( \Rightarrow(\sin x+1)=2(1-\sin x)(1+\sin x) \)
\( \Rightarrow(1+\sin x)[2(1-\sin x)-1]=0 \)
\( \Rightarrow 2(1-\sin x)-1=0\quad \ldots[\sin x \neq-1 \text { otherwise } \cos x=0 \text { and } \)\( \tan x, \sec x \text { will be undefined}] \)
\( \Rightarrow \sin x=\frac{1}{2}\)
\( \Rightarrow x=\frac{\pi}{6}, \frac{5 \pi}{6} \text { in }(0,2 \pi)\)
Now, \(\tan x+\sec x=2 \cos x\)
|(\Rightarrow \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x \)
\( \Rightarrow(\sin x+1)=2 \cos ^2 x \)
\( \Rightarrow(\sin x+1)=2\left(1-\sin ^2 x\right) \)
\( \Rightarrow(\sin x+1)=2(1-\sin x)(1+\sin x) \)
\( \Rightarrow(1+\sin x)[2(1-\sin x)-1]=0 \)
\( \Rightarrow 2(1-\sin x)-1=0\quad \ldots[\sin x \neq-1 \text { otherwise } \cos x=0 \text { and } \)\( \tan x, \sec x \text { will be undefined}] \)
\( \Rightarrow \sin x=\frac{1}{2}\)
\( \Rightarrow x=\frac{\pi}{6}, \frac{5 \pi}{6} \text { in }(0,2 \pi)\)
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