Q.35. If general solution of cos^{2}\theta -2sin\theta +\frac{1}{4}=0 is \theta =\frac{n\pi }{A}+(-1)^{n}\frac{\pi }{B}, n\epsilon Z , then A+B has the
A. 7
B. 6
C. 1
D. -7
Answer :- A. 7
Explanation :-
\cos^2\theta - 2\sin\theta + \frac{1}{4} = 0 (1-\sin^2\theta) - 2\sin\theta + \frac{1}{4} = 0 \sin^2\theta + 2\sin\theta - \frac{5}{4} = 0 4\sin^2\theta + 8\sin\theta - 5 = 0 4\sin^2\theta + 10\sin\theta - 2\sin\theta - 5 = 0 2\sin\theta(2\sin\theta+5) - 1(2\sin\theta+5) = 0 (2\sin\theta-1)(2\sin\theta+5) = 0\sin\theta = \frac{1}{2} or \sin\theta = \frac{-5}{2}
But \sin\theta = \frac{-5}{2} is not possible as \sin\theta \in [-1,1] for all values of \theta.
\sin\theta = \frac{1}{2} \sin\theta = \sin\frac{\pi}{6} \theta = \frac{n\pi}{1} + (-1)^n\frac{\pi}{6}A=1 and B=6
\Rightarrow A+B=7