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

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

The number of values of θ ∈ [(- 3π)/(2), (4π)/(3)] which satisfies...

The number of values of θ∈\theta \in\left[\dfrac{- 3\pi}{2}, \dfrac{4\pi}{3}\right]whichsatisfiesthesystemofequationswhich satisfies the system of equations2\sin^{2}\theta + \sin^{2}2\theta = 2 and sin⁡2θ+cos⁡2θ=tanθ\sin 2\theta + \cos 2\theta = tan\theta is

Options

  1. A22
  2. B44
  3. C66
  4. D

    88

Correct answer

C. 66

Step-by-step solution

First equation is 4sin⁡2θcos2θ=2cos⁡2θ4\sin^{2}\theta{cos}^{2}\theta = 2\cos^{2}\theta

⇒cos⁡2θ=0⇒ \cos^{2} \theta = 0 or sin⁡2θ=12\sin^{2}\theta = \dfrac{1}{2}=(12)2=sin⁡2π4= {\left(\dfrac{1}{\sqrt{2}}\right)}^{2} = \sin^{2}\dfrac{\pi}{4}

⇒θ=(2n+1)π2,n∈I⇒ \theta = \left(2n + 1\right)\dfrac{\pi}{2}, n \in I or θ=nπ±π4,n∈I\theta = n\pi \pm \dfrac{\pi}{4}, n \in I

Second equation is not satisfied by θ=(2n+1)π2,n∈I\theta = \left(2n + 1\right)\dfrac{\pi}{2}, n \in I but satisfied by θ=nπ±π4,n∈I\theta = n\pi \pm \dfrac{\pi}{4}, n \in I

So θ=n \pi±π4,n∈I\theta = \text{n \pi} \pm \dfrac{\pi}{4}, n \in I

∴∴ In [−3π2,4π3][ \dfrac{- 3\pi}{2}, \dfrac{4\pi}{3} ], the values of θ\theta are

−5π4,−3π4,−π4,π4,3π4,5π4\dfrac{- 5\pi}{4}, \dfrac{- 3\pi}{4}, \dfrac{- \pi}{4}, \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}

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