Trigonometric Equations

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Question 1 of 15

Solve the equation (2\sin^2\theta + \sqrt{3}\sin\theta - 1 = 0) for (0 \le \theta \le 2\pi).

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\(\theta = \frac{\pi}{3}, \frac{5\pi}{3}\)

\(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)

\(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)

\(\theta = \frac{\pi}{2}, \frac{3\pi}{2}\)

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