Question

Statement-I: In the interval \( [0, 2\pi] \), the number of common solutions of the equations

\[ 2\sin^2\theta - \cos 2\theta = 0 \]

and

\[ 2\cos^2\theta - 3\sin\theta = 0 \]

is two.

Statement-II: The number of solutions of

\[ 2\cos^2\theta - 3\sin\theta = 0 \]

in \( [0, \pi] \) is two.

• A. Statement-I and Statement-II are both true
• B. Statement-I is true, Statement-II is false
• C. Statement-I is false, Statement-II is true
• D. Statement-I and Statement-II are both false

Hint:

When comparing solutions of trigonometric equations, use standard identities and substitution to reduce both equations to a common trigonometric function for easier comparison.

Answer 1

The Correct Option is A

Start with the two equations: **Equation 1:** \[ 2\sin^2\theta - \cos 2\theta = 0 \] Use identity: \( \cos 2\theta = 1 - 2\sin^2\theta \), so: \[ 2\sin^2\theta - (1 - 2\sin^2\theta) = 0 \Rightarrow 2\sin^2\theta - 1 + 2\sin^2\theta = 0 \Rightarrow 4\sin^2\theta = 1 \Rightarrow \sin\theta = \pm\frac{1}{2} \] So from \( \sin\theta = \pm \frac{1}{2} \), in \( [0, 2\pi] \), we get 4 values of \( \theta \): \( \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \) **Equation 2:** \[ 2\cos^2\theta - 3\sin\theta = 0 \Rightarrow 2(1 - \sin^2\theta) - 3\sin\theta = 0 \Rightarrow 2 - 2\sin^2\theta - 3\sin\theta = 0 \Rightarrow 2\sin^2\theta + 3\sin\theta - 2 = 0 \] Let \( y = \sin\theta \), solve: \[ 2y^2 + 3y - 2 = 0 \Rightarrow y = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm \sqrt{25}}{4} = \frac{-3 \pm 5}{4} \Rightarrow y = \frac{1}{2},\ -2 \] Only \( \sin\theta = \frac{1}{2} \) is valid (as \( \sin\theta \in [-1, 1] \)). \[ \text{So from this, the possible values are } \frac{\pi}{6}, \frac{5\pi}{6} \in [0, \pi].
\text{Hence, there are 2 solutions in } [0, \pi] \Rightarrow Statement-II is true. \] Now check how many values are common in both equations in \( [0, 2\pi] \): From Equation 1: \( \sin\theta = \pm \frac{1}{2} \Rightarrow \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \)
From Equation 2: \( \sin\theta = \frac{1}{2} \Rightarrow \frac{\pi}{6}, \frac{5\pi}{6} \) **Common values:** \( \frac{\pi}{6}, \frac{5\pi}{6} \Rightarrow 2 \) values Statement-I is also true. Hence, both Statement-I and Statement-II are true.