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JEE Main 2002
Exam
JEE Main
Year
2002
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2002 · Mathematics · Trigonometric Equations

The number of solution of x+ x=2 x in [0,2 π) is

The number of solution of tan⁡x+sec⁡x=2cos⁡x\tan x+\sec x=2 \cos x in [0,2π)[0,2 \pi) is

Options

  1. A2
  2. B3
  3. C0
  4. D1

Correct answer

B. 3

Step-by-step solution

The given equation is tan⁡x+sec⁡x=2cos⁡x⇒sin⁡x+1=2cos⁡2x\tan x+\sec x=2 \cos x \Rightarrow \sin x+1=2 \cos ^2 x⇒sin⁡x+1=2(1−sin⁡2x)⇒2sin⁡2x+sin⁡x−1=0⇒(2sin⁡x−1)(sin⁡x+1)=0⇒sin⁡x=12,−1⇒x=30∘,150∘,270∘\begin{aligned}& \Rightarrow \sin x+1=2 \left(1-\sin ^2 x\right) \Rightarrow 2 \sin ^2 x+\sin x-1=0 \\ & \Rightarrow(2 \sin x-1)(\sin x+1)=0 \Rightarrow \sin x=\frac{1}{2},-1 \Rightarrow x=30^{{}^{\circ}}, 150^{{}^{\circ}}, 270^{{}^{\circ}} \end{aligned}$
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