$1 - \cos^2\theta - 2\cos\theta + \frac{1}{4} = 0$
$4\cos^2\theta + 8\cos\theta - 5 = 0$
$(2\cos\theta - 1)(2\cos\theta + 5) = 0$
$\cos\theta = \frac{1}{2}$ (rejecting $-5/2$)
$$\theta = 2n\pi \pm \frac{\pi}{3}$$
Back to JEE Main 2023 Trigonometry
JEE Main 2023 — Mathematics Trigonometry
hard
mcq
2023
Official previous-year question
Verified 30 May 2026.
Question
The general solution of $\sin^2\theta - 2\cos\theta + \frac{1}{4} = 0$ is:
Options
- A
$\theta = 2n\pi \pm \frac{\pi}{3}$
- B
$\theta = 2n\pi \pm \frac{\pi}{6}$
- C
$\theta = n\pi \pm \frac{\pi}{3}$
- D
$\theta = n\pi + (-1)^n\frac{\pi}{6}$
Solution
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