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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: On the domain [−π,π][-\pi, \pi] the equation 4sin⁡3x+2sin⁡2x−2sin⁡x−1=04 \sin ^3 x+2 \sin ^2 x-2 \sin x-1=0 possess

Options

  1. AOnly one real root
  2. BThree real roots
  3. CFour real roots
  4. DSix real roots

Correct answer

D. Six real roots

Step-by-step solution

4sin⁡3x+2sin⁡2x−2sin⁡x−1=(2sin⁡x+1)(2sin⁡2x−1)=0∴sin⁡x=−12,±12∴4 \sin ^3 x+2 \sin ^2 x-2 \sin x-1=(2 \sin x+1)\left(2 \sin ^2 x-1\right)=0\therefore \quad \sin x=-\frac{1}{2}, \pm \frac{1}{\sqrt{2}}\therefore \quad There are 6 solutions.
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