UPSC CDS 1 2023 Math Paper

Concept:Use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to rewrite the equation as a quadratic in $\sin \theta$ and then solve.Explanation:Start with $2 \cos^2 \theta + \sin \theta - 2 = 0$.Substitute $\cos^2 \theta = 1 - \sin^2 \theta$: $2(1 - \sin^2 \theta) + \sin \theta - 2 = 0$.Expand and simplify: $2 - 2\sin^2 \theta + \sin \theta - 2 = 0$, which becomes $-2\sin^2 \theta + \sin \theta = 0$.Factor out $\sin \theta$: $\sin \theta (-2\sin \theta + 1) = 0$.So either $\sin \theta = 0$ or $\sin \theta = \frac{1}{2}$.Given $0 < \theta \leq \frac{\pi}{2}$, $\sin \theta = 0$ would imply $\theta = 0$ (not allowed because $\theta > 0$) or $\theta = \pi$ (outside range). Therefore discard $\sin \theta = 0$.Thus $\sin \theta = \frac{1}{2}$.In the interval $(0,\frac{\pi}{2}]$, $\sin \theta = \frac{1}{2}$ gives $\theta = \frac{\pi}{6}$.Verification: $\theta = \frac{\pi}{6}$ satisfies $2\cos^2\left(\frac{\pi}{6}\right) + \sin\left(\frac{\pi}{6}\right) - 2 = 2\cdot\left(\frac{\sqrt{3}}{2}\right)^2 + \frac{1}{2} - 2 = \frac{3}{2} + \frac{1}{2} - 2 = 0$.Hence the correct value is $\theta = \frac{\pi}{6}$.Answer:$\theta = \frac{\pi}{6}$ (Option A).