Question

Number of solutions of the equation $\sin^2\theta + 2\cos^2\theta - \sqrt{3}\sin\theta\cos\theta = 2$ lying in the interval $(-\pi, \pi)$ is

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

For trigonometric equations, be extremely careful about the specified interval (whether it's open or closed). A small difference in the interval can change the number of solutions. When an answer doesn't match, re-read the question and consider if the interval notation might be interpreted differently (e.g., a typo from `[` to `(`). Using double angle formulas is often a robust way to solve equations involving squared terms.

Answer 1

The Correct Option is C

Step 1: Simplify the Equation: \[ \sin^2\theta + 2\cos^2\theta - \sqrt{3}\sin\theta\cos\theta = 2 \] Split \( 2\cos^2\theta \) into \( \cos^2\theta + \cos^2\theta \): \[ (\sin^2\theta + \cos^2\theta) + \cos^2\theta - \sqrt{3}\sin\theta\cos\theta = 2 \] \[ 1 + \cos^2\theta - \sqrt{3}\sin\theta\cos\theta = 2 \] \[ \cos^2\theta - \sqrt{3}\sin\theta\cos\theta = 1 \]
Step 2: Solve the Simplified Equation: Rearrange to get: \[ \cos^2\theta - 1 = \sqrt{3}\sin\theta\cos\theta \] \[ -\sin^2\theta = \sqrt{3}\sin\theta\cos\theta \] \[ \sin^2\theta + \sqrt{3}\sin\theta\cos\theta = 0 \] \[ \sin\theta (\sin\theta + \sqrt{3}\cos\theta) = 0 \]
Step 3: Find Solutions in \( (-\pi, \pi) \): Case 1: \( \sin\theta = 0 \) In \( (-\pi, \pi) \), the only solution is \( \theta = 0 \). (Endpoints are excluded). Case 2: \( \sin\theta + \sqrt{3}\cos\theta = 0 \) \[ \tan\theta = -\sqrt{3} \] In \( (-\pi, \pi) \), tangent is negative in the 2nd and 4th quadrants. - 4th Quadrant: \( \theta = -\frac{\pi}{3} \) - 2nd Quadrant: \( \theta = \frac{2\pi}{3} \) Total solutions are \( \{ -\frac{\pi}{3}, 0, \frac{2\pi}{3} \} \). Count = 3.