Q.30. 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.{\frac{2\pi}{3}, \frac{4\pi}{3}}
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}
Since \cos\theta = \frac{-5}{4} is not possible (as cosine values range between -1 and 1), we consider:
\cos\theta = -\frac{1}{2}From unit circle properties, \cos\theta = -\frac{1}{2} at \theta = \frac{2\pi}{3} and \theta = \frac{4\pi}{3}.
Thus, the solution set is {\frac{2\pi}{3}, \frac{4\pi}{3}}.