Trigonometry – Trigonometric Equations – 22 January 2025 (Shift 2)

Solution:

Eq 1: $2\sin^2\theta = \cos2\theta$
$2\sin^2\theta = 1 – 2\sin^2\theta \Rightarrow 4\sin^2\theta = 1 \Rightarrow \sin\theta = \pm \frac{1}{2}$.

Eq 2: $2\cos^2\theta = 3\sin\theta$
$2(1-\sin^2\theta) = 3\sin\theta \Rightarrow 2 – 2\sin^2\theta – 3\sin\theta = 0$
$2\sin^2\theta + 3\sin\theta – 2 = 0$
$(2\sin\theta – 1)(\sin\theta + 2) = 0$.
Since $\sin\theta \neq -2$, we have $\sin\theta = \frac{1}{2}$.

Common solution: $\sin\theta = \frac{1}{2}$.
In $[0, 2\pi]$, solutions are $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$.
Sum $= \frac{\pi}{6} + \frac{5\pi}{6} = \pi$.

Ans. (4)