If the general solution of cos ^2 theta - 2 s | vedclass

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(A) Given equation: $\cos ^2 	heta - 2 \sin 	heta + \frac{1}{4} = 0$ Substitute $\cos ^2 	heta = 1 - \sin ^2 	heta$: $(1 - \sin ^2 	heta) - 2 \si

Vedclass · Accepted Answer

$7$

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(A) Given equation: $\cos ^2 	heta - 2 \sin 	heta + \frac{1}{4} = 0$ Substitute $\cos ^2 	heta = 1 - \sin ^2 	heta$: $(1 - \sin ^2 	heta) - 2 \sin 	heta + \frac{1}{4} = 0$ $\sin ^2 	heta + 2 \sin 	heta - \frac{5}{4} = 0$ Multiply by $4$: $4 \sin ^2 	heta + 8 \sin 	heta - 5 = 0$ Factorize the quadratic: $4 \sin ^2 	heta + 10 \sin 	heta - 2 \sin 	heta - 5 = 0$ $2 \sin 	heta(2 \sin 	heta + 5) - 1(2 \sin 	heta + 5) = 0$ $(2 \sin 	heta - 1)(2 \sin 	heta + 5) = 0$ So,$\sin 	heta = \frac{1}{2}$ or $\sin 	heta = -\frac{5}{2}$. Since $-1 \le \sin 	heta \le 1$,$\sin 	heta = -\frac{5}{2}$ is impossible. Thus,$\sin 	heta = \frac{1}{2} = \sin \frac{\pi}{6}$. The general solution is $	heta = n \pi + (-1)^n \frac{\pi}{6}$. Comparing with $	heta = \frac{n \pi}{A} + (-1)^n \frac{\pi}{B}$,we get $A = 1$ and $B = 6$. Therefore,$A + B = 1 + 6 = 7$.

If the general solution of $\cos ^2 \theta - 2 \sin \theta + \frac{1}{4} = 0$ is $\theta = \frac{n \pi}{A} + (-1)^{n} \frac{\pi}{B}, n \in Z$,then $A + B$ has the value

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