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

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

The solution set of the equation 4sin θ cos θ - 2cos θ - 2 3 sin θ...

The solution set of the equation 4sinθcosθ−2cosθ−23sinθ+3=04sin \theta cos \theta - 2cos \theta - 2\sqrt{3}sin \theta + \sqrt{3} = 0 in the interval (0,2π)\left(0, 2\pi\right) is

Options

  1. A{3π4,7π4}\left\{\dfrac{3\pi}{4}, \dfrac{7\pi}{4}\right\}
  2. B{π3,5π3}\left\{\dfrac{\pi}{3}, \dfrac{5\pi}{3}\right\}
  3. C{3π4,π,π3,5π3}\left\{\dfrac{3\pi}{4}, \pi, \dfrac{\pi}{3}, \dfrac{5\pi}{3}\right\}
  4. D{π6,5π6,11π6}\left\{\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{11\pi}{6}\right\}

Correct answer

D. {π6,5π6,11π6}\left\{\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{11\pi}{6}\right\}

Step-by-step solution

4sin \theta cos \theta - 2cos \theta - 2\sqrt{3}sin \theta + \sqrt{3} = 0<br/><br />2cos \theta\left(2sin \theta - 1\right) - \sqrt{3}\left(2sin \theta - 1\right) = 0<br/>(2cosθ−3)(2sinθ−1)=0<br />\left(2cos \theta - \sqrt{3}\right)\left(2sin \theta - 1\right) = 0
⇒sinθ=12,cosθ=32⇒ sin \theta = \dfrac{1}{2}, cos \theta = \dfrac{\sqrt{3}}{2}
θ=π6,5π6,11π6\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{11\pi}{6}
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