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

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

cose c ^ 2 θ( ^ 2 θ - 3cos θ + 2) ≥ 1, if θ belongs to

cosec2θ(cos⁡2θ−3cosθ+2)≥1,cose{c}^{2}\theta\left(\cos^{2}\theta - 3cos \theta + 2\right) \ge 1, if θ\theta belongs to

Options

  1. A(0,π3)\left(0, \dfrac{\pi}{3}\right)
  2. B(π2,π)\left(\dfrac{\pi}{2}, \pi\right)
  3. C(π3,π2)\left(\dfrac{\pi}{3}, \dfrac{\pi}{2}\right)
  4. D(0,π4)\left(0, \dfrac{\pi}{4}\right)

Correct answer

C. (π3,π2)\left(\dfrac{\pi}{3}, \dfrac{\pi}{2}\right)

Step-by-step solution

\cos^{2} \theta - 3cos \theta + 2 \ge \dfrac{1}{{cosec}^{2}\theta} = \sin^{2} \theta<br/><br />\cos^{2} \theta - 3cos \theta + 2 \ge 1 - \cos^{2} \theta<br/><br />2\cos^{2} \theta - 3cos \theta + 1 \ge 0<br/>(2cosθ−1)(cosθ−1)≥0<br />\left(2cos \theta - 1\right)\left(cos \theta - 1\right) \ge 0
⇒cosθ≤12⇒ cos \theta \le \dfrac{1}{2} or cosθ≥1cos \theta \ge 1
So, θ∈(π3,π2)\theta \in \left(\dfrac{\pi}{3}, \dfrac{\pi}{2}\right) of the given intervals.
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