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JEE Advanced 2016Paper 1
Exam
JEE Advanced
Year
2016
Paper
Paper 1
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Advanced 2016 · Paper 1 · Mathematics · Trigonometric Equations

Let S = \ x ∈ (- π, π): x ≠ 0, ± (π)/(2)\ . The sum of all distinct...

Let S={x∈(−π,π):x≠0,±π2}S = \left\{x \in \left(- \pi, \pi\right): x \ne 0, \pm \dfrac{\pi}{2}\right\} . The sum of all distinct solution of the equation 3secx+cosecx+2(tanx−cotx)=0\sqrt{3}sec x + cosecx + 2\left(tan x - cot x\right) = 0, in the set S is equal to

Options

  1. A−7π9- \dfrac{7\pi}{9}
  2. B−2π9- \dfrac{2\pi}{9}
  3. C00
  4. D5π9\dfrac{5\pi}{9}

Correct answer

C. 00

Step-by-step solution

3cos⁡x+1sin⁡x+2(sin⁡xcos⁡x−cos⁡xsin⁡x)=0\dfrac{\sqrt{3}}{\cos x} + \dfrac{1}{\sin x} + 2\left(\dfrac{\sin x}{\cos x} - \dfrac{\cos x}{\sin x}\right) = 0
3sinx+cosx=2cos⁡2x\sqrt{3}sin x + cos x = 2\cos 2x
⇒⇒ cos⁡2x=cos⁡(x−π3)\cos 2x = \cos \left(x - \dfrac{\pi}{3}\right)
⇒2x=2nπ±(x−π3)⇒ 2x = 2n\pi \pm \left(x - \dfrac{\pi}{3}\right),\left\{2x=2nπ+x−π32x=2nπ−x+π3\begin{array}{ll}2x = 2n\pi + x - \dfrac{\pi}{3} \\ 2x = 2n\pi - x + \dfrac{\pi}{3}\end{array}\right)
⇒x=(6n−1)π3⇒ x = \left(6n - 1\right)\dfrac{\pi}{3} or \left(6n + 1\right)\dfrac{\pi}{9}<br/><br />⇒ x = - \dfrac{\pi}{3}, \dfrac{\pi}{9}, \dfrac{7\pi}{9}andand- \dfrac{5\pi}{9}in(−π,π)in \left(- \pi, \pi\right)
⇒sum=0⇒ sum = 0
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