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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

The equation x^3-(3)/(4) x=- 3 8 is satisfied by

The equation x3−34x=−38x^3-\frac{3}{4} x=-\frac{\sqrt{3}}{8} is satisfied by

Options

  1. Ax=cos⁡(5π18)x=\cos \left(\frac{5 \pi}{18}\right)
  2. Bx=cos⁡(7π18)x=\cos \left(\frac{7 \pi}{18}\right)
  3. Cx=cos⁡(23π18)x=\cos \left(\frac{23 \pi}{18}\right)
  4. Dx=cos⁡(17π18)x=\cos \left(\frac{17 \pi}{18}\right)

Correct answer

D. x=cos⁡(17π18)x=\cos \left(\frac{17 \pi}{18}\right)

Step-by-step solution

Let x=cos⁡θx=\cos \theta4cos⁡3θ−3cos⁡θ=−32cos⁡3θ=cos⁡5π6\begin{aligned} & 4 \cos ^3 \theta-3 \cos \theta=-\frac{\sqrt{3}}{2} \\ & \cos 3 \theta=\cos \frac{5 \pi}{6}\end{aligned}$\begin{aligned} & \Rightarrow 3 \theta=2 \mathrm{n} \pi \pm \frac{5 \pi}{6} \\ & \Rightarrow \quad \theta=\frac{2 \mathrm{n} \pi}{3} \pm \frac{5 \pi}{18}\end{aligned} put n=0,θ=5π18n=1,θ=2π3+5π18=17π18;θ=2π3−5π18=7π18\begin{aligned} & \text { put } \mathrm{n}=0, \theta=\frac{5 \pi}{18} \\ & \mathrm{n}=1, \theta=\frac{2 \pi}{3}+\frac{5 \pi}{18}=\frac{17 \pi}{18} ; \theta=\frac{2 \pi}{3}-\frac{5 \pi}{18}=\frac{7 \pi}{18}\end{aligned}$
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