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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

Let 0 < θ _ 1 < θ _ 2 < θ _ 3 < denotes the positive solutions of...

Let 0<θ1<θ2<θ3<…0 < {\theta}_{1} < {\theta}_{2} < {\theta}_{3} < \ldots denotes the positive solutions of the equation 3+3cosθ=2sin⁡2θ3 + 3cos \theta = 2\sin^{2} \theta. If θ3+θ7=aπ{\theta}_{3} + {\theta}_{7} = a\pi, where aa is an integer, then the value of aa is equal to

Options

  1. A66
  2. B77
  3. C88
  4. D44

Correct answer

A. 66

Step-by-step solution

3+3cosθ=2sin⁡2θ=2−2cos⁡2θ3 + 3cos \theta = 2\sin^{2} \theta = 2 - 2\cos^{2} \theta.

⇒2cos⁡2θ+3cosθ+1=0⇒ 2\cos^{2} \theta + 3cos \theta + 1 = 0

⇒cosθ=−1,−12⇒ cos \theta = - 1, - \dfrac{1}{2}

⇒θ=⇒ \theta =\left(2n + 1\right)\pioror2n\pi \pm \dfrac{2\pi}{3}<br/><br/><br /><br />⇒ \theta = \dfrac{2\pi}{3}, \pi, \dfrac{4\pi}{3}, \dfrac{8\pi}{3}, 3\pi, \dfrac{10\pi}{3}, \dfrac{14\pi}{3}, 5\pi<br/><br/>i.e.<br /><br />i.e.{\theta}_{3} + {\theta}_{7} = \dfrac{4\pi}{3} + \dfrac{14\pi}{3} = \dfrac{18\pi}{3} = 6\pi<br/><br/><br /><br />⇒ a = 6
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