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

\(\cos ^2 \theta-2 \sin \theta+\frac{1}{4}=0 \)
\( \therefore \left(1-\sin ^2 \theta\right)-2 \sin \theta+\frac{1}{4}=0 \)
\( \therefore \sin ^2 \theta+2 \sin \theta-\frac{5}{4}=0 \)
\( \therefore 4 \sin ^2 \theta+8 \sin \theta-5=0 \)
\( \therefore 4 \sin ^2 \theta+10 \sin \theta-2 \sin \theta-5=0 \)
\( \therefore 2 \sin \theta(2 \sin \theta+5)-1(2 \sin \theta+5)=0 \)
\( \therefore (2 \sin \theta-1)(2 \sin \theta+5)=0 \)
\( \therefore \sin \theta=\frac{1}{2} \text { or } \sin \theta=\frac{-5}{2} \)
But \(\sin \theta=\frac{-5}{2}\) is not possible as \(\sin \theta \in[-1,1]\)
\( \text {for all values of } \theta . \)
\( \therefore \sin \theta=\frac{1}{2} \)
\( \therefore \sin \theta=\sin \frac{\pi}{6} \)
\( \therefore \theta=\frac{n \pi}{1}+(-1)^{\mathrm{n}} \frac{\pi}{6} \)
\( \therefore \text {A} \Rightarrow \text {A and } B=6 \)
\( \Rightarrow \text {B}=7\)