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

Question

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

Aicharya · Accepted Answer

$\begin{array}{ll} & \cos ^2 	heta-2 \sin 	heta+\frac{1}{4}=0 \ 	herefore \quad & \left(1-\sin ^2 	hetaight)-2 \sin 	heta+\frac{1}{4}=0 \ 	herefore \quad & \sin ^2 	heta+2 \sin 	heta-\frac{5}{4}=0 \ 	herefore \quad & 4 \sin ^2 	heta+8 \sin 	heta-5=0 \ 	herefore \quad & 4 \sin ^2 	heta+10 \sin 	heta-2 \sin 	heta-5=0 \ 	herefore \quad & 2 \sin 	heta(2 \sin 	heta+5)-1(2 \sin 	heta+5)=0 \ 	herefore \quad & (2 \sin 	heta-1)(2 \sin 	heta+5)=0 \ 	herefore \quad & \sin 	heta=\frac{1}{2} 	ext { or } \sin 	heta=\frac{-5}{2} \ & 	ext { But } \sin 	heta=\frac{-5}{2} 	ext { is not possible as sin } 	heta \in[-1,1] \ & 	ext { for all values of } 	heta . \ 	herefore \quad & \sin 	heta=\frac{1}{2} \ 	herefore \quad & \sin 	heta=\sin \frac{\pi}{6} \ 	herefore \quad & 	heta=\frac{n \pi}{1}+(-1)^{\mathrm{n}} \frac{\pi}{6} \ 	ext{A} & = 6	ext { and } B=8 \ & 	herefore, A+B=1+6=7\end{array}$

If general solution of $\cos ^2 \theta-2 \sin \theta+\frac{1}{4}=0$ is $\theta=\frac{\mathrm{n}…

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