Q.31. The solution set of 8\cos^2\theta+14\cos\theta+5=0, in the interval [0, 2\pi], is
A. \{\frac{\pi}{3}, \frac{2\pi}{3}\}
B. \{\frac{\pi}{3}, \frac{4\pi}{3}\}
C. \{\frac{2\pi}{3}, \frac{4\pi}{3}\}
D. \{\frac{2\pi}{3}, \frac{5\pi}{3}\}
Answer: C
Explanation:
8\cos^2\theta+14\cos\theta+5=0 8\cos^2\theta + 10\cos\theta + 4\cos\theta + 5 = 0 2\cos\theta(4\cos\theta + 5) + 1(4\cos\theta + 5) = 0 (2\cos\theta + 1)(4\cos\theta + 5) = 0\cos\theta = -\frac{1}{2} or \cos\theta = -\frac{5}{4}
At \cos\theta = \frac{-5}{4}, it is not possible as \cos\theta \in [-1, 1] for all values of \theta.
\cos\theta = \frac{-1}{2} \theta \in \left\{\frac{2\pi}{3}, \frac{4\pi}{3} \right\}