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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

Paragraph: Let a, b, c, d ∈ R . Then the cubic equation of the type...

Paragraph: Let a,b,c,d∈Ra, b, c, d \in R. Then the cubic equation of the type ax3+bx2+cx+d=0a x^3+b x^2+c x+d=0 has either one root real or all three roots are real. But in case of trigonometric equations of the type asin⁡3x+bsin⁡2x+csin⁡x+d=0a \sin ^3 x+b \sin ^2 x+c \sin x+d=0 can possess several solutions depending upon the domain of xx. To solve an equation of the type acos⁡θ+bsin⁡θ=ca \cos \theta+b \sin \theta=c. The equation can be written as cos⁡(θ−α)=c/(a2+b2)\cos (\theta-\alpha)=c / \sqrt{\left(a^2+b^2\right)}. The solution is θ=2nπ+α±β\theta=2 \mathrm{n} \pi+\alpha \pm \beta, where tan α=b/a,cos⁡β=c/(a2+b2)\alpha=b / a, \cos\beta=c / \sqrt{\left(a^2+b^2\right)}Question: ∣tan⁡x∣=tan⁡x+1cos⁡x(0≤x≤2π)|\tan x|=\tan x+\frac{1}{\cos x}(0 \leq x \leq 2 \pi) has

Options

  1. ANo solution
  2. BOne solution
  3. CTwo solutions
  4. DThree solutions

Correct answer

B. One solution

Step-by-step solution

(i) When tan⁡x≥0\tan x \geq 0, then the equation becomes tan⁡x=tan⁡x+1cos⁡x\tan x=\tan x+\frac{1}{\cos x} i.e 1cos⁡x=0\frac{1}{\cos x}=0 (not possible) (ii) When tan⁡x<0\tan x \lt 0, then the equation becomes −tan⁡x=tan⁡x+1cos⁡x-\tan x=\tan x+\frac{1}{\cos x} i.e sin⁡x=−12∴x=11π6\sin x=-\frac{1}{2}\therefore \quad x=\frac{11 \pi}{6} is the only solution.
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