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JEE Main 201322 AprilOnline
Exam
JEE Main
Year
2013
Date
22 April
Mode
Online
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2013 · 22 April · Online · Mathematics · Trigonometric Equations

Statement-1: The number of common solutions of the trigonometric...

Statement-1: The number of common solutions of the trigonometric equations 2sin⁡2θ2 \sin ^2 \theta −cos⁡2θ=0-\cos 2 \theta=0 and 2cos⁡2θ−3sin⁡θ=02 \cos ^2 \theta-3 \sin \theta=0 in the interval [0,2π][0,2 \pi] is two. Statement-2: The number of solutions of the equation, 2cos⁡2θ−3sin⁡θ=02 \cos ^2 \theta-3 \sin \theta=0 in the interval [0,π][0, \pi] is two.

Options

  1. AStatement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1.
  2. BStatement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.
  3. CStatement-1 is false; Statement-2 is true.
  4. DStatement-1 is true; Statement-2 is false.

Correct answer

B. Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.

Step-by-step solution

2sin⁡2θ−cos⁡2θ=0⇒2sin⁡2θ−(1−2sin⁡2θ)=0⇒2sin⁡2θ−1+2sin⁡2θ=0⇒4sin⁡2θ=1⇒sin⁡θ=±12∴θ=π4,3π4,5π4,7π4,θ∈[0,2π]∴θ=π6,5π6,7π6,11π62 \sin ^2 \theta-\cos 2 \theta=0\Rightarrow 2 \sin ^2 \theta-\left(1-2 \sin ^2 \theta\right)=0\Rightarrow 2 \sin ^2 \theta-1+2 \sin ^2 \theta=0\Rightarrow 4 \sin ^2 \theta=1 \Rightarrow \sin \theta=\pm \frac{1}{2}\therefore \theta=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}, \theta \in[0,2 \pi]\therefore \theta=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6} Now 2cos⁡2θ−3sin⁡θ=0⇒2(1−sin⁡2θ)−3sin⁡θ=0⇒−2sin⁡2θ−3sin⁡θ+2=0⇒−2sin⁡2θ−4sin⁡θ+sin⁡θ+2=0⇒2sin⁡2θ−sin⁡θ+4sin⁡θ−2=0⇒sin⁡θ(2sin⁡θ−1)+2(2sin⁡θ−1)=0⇒sin⁡θ=12,−2\begin{aligned}& \text{ Now }2 \cos ^2 \theta-3 \sin \theta=0 \\ \Rightarrow & 2 \left(1-\sin ^2 \theta\right)-3 \sin \theta=0 \\ \Rightarrow & -2 \sin ^2 \theta-3 \sin \theta+2=0 \\ \Rightarrow & -2 \sin ^2 \theta-4 \sin \theta+\sin \theta+2=0 \\ \Rightarrow & 2 \sin ^2 \theta-\sin \theta+4 \sin \theta-2=0 \\ \Rightarrow & \sin \theta(2 \sin \theta-1)+2(2 \sin \theta-1)=0 \\ \Rightarrow & \sin \theta=\frac{1}{2},-2 \end{aligned}ButBut\sin \theta=-2,isnotpossible∴sin⁡, is not possible \therefore \quad \sin\theta=12\frac{1}{2}, \Rightarrow \theta=π6\frac{\pi}{6}, 5π6\frac{5 \pi}{6}Hence, there are two common solution, there each of the statement-1 and 2 are true but statement- 2 is not a correct explanation for statement-1.
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