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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: In the interval [−π/4,π/2][-\pi / 4, \pi / 2], the equation, cos⁡4x+10tan⁡x1+tan⁡2x=3\cos 4 x+\frac{10 \tan x}{1+\tan ^2 x}=3 has

Options

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

Correct answer

C. Two solutions

Step-by-step solution

3=cos⁡4x+10tan⁡x1+tan⁡2x=cos⁡4x+5sin⁡2x3=\cos 4 x+\frac{10 \tan x}{1+\tan ^2 x}=\cos 4 x+5 \sin 2 x i.e 3=1−2sin⁡22x+5sin⁡2x3=1-2 \sin ^2 2 x+5 \sin 2 x i.e sin⁡2x=12∴2x=π6,5π6\sin 2 x=\frac{1}{2}\therefore \quad 2 x=\frac{\pi}{6}, \frac{5 \pi}{6} Thus there are two solutions.
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