Question

Number of solutions of the equation \[ 2 \sin^2 \theta - 3 \cos^2 \theta = \sin \theta \cos \theta \] in the interval \((- \pi, \pi)\) is:

• A. 2
• B. 4
• C. 3
• D. 1

Hint:

Convert trigonometric equations to tangent form and solve quadratic equations for solutions.

Answer 1

The Correct Option is B

Step 1: Express equation in terms of \(\tan \theta\)
Rewrite: \[ 2 \sin^2 \theta - 3 \cos^2 \theta = \sin \theta \cos \theta \] Divide both sides by \(\cos^2 \theta\) (where \(\cos \theta \neq 0\)): \[ 2 \tan^2 \theta - 3 = \tan \theta \] Step 2: Rearrange to quadratic in \(\tan \theta\)
\[ 2 \tan^2 \theta - \tan \theta - 3 = 0 \] Step 3: Solve quadratic
\[ \tan \theta = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm 5}{4} \] So, \[ \tan \theta = \frac{6}{4} = \frac{3}{2}, \quad \text{or} \quad \tan \theta = \frac{-4}{4} = -1 \] Step 4: Find solutions in \((- \pi, \pi)\)
Each value corresponds to two solutions in \((- \pi, \pi)\), total 4 solutions.