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(B) For set $A$: $\cos^2 x = \cos^2 \frac{\pi}{6} \implies \cos 2x = \cos 2(\frac{\pi}{6}) = \cos \frac{\pi}{3}$.This gives $2x = 2n\pi \pm \frac{\pi}

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$\{x \in R \mid x = 2n\pi \pm \frac{\pi}{3}, 2n\pi \pm \frac{2\pi}{3}, n \in Z\}$

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(B) For set $A$: $\cos^2 x = \cos^2 \frac{\pi}{6} \implies \cos 2x = \cos 2(\frac{\pi}{6}) = \cos \frac{\pi}{3}$.This gives $2x = 2n\pi \pm \frac{\pi}{3} \implies x = n\pi \pm \frac{\pi}{6}$.Thus,$A = \{n\pi \pm \frac{\pi}{6} \mid n \in Z\}$.For set $B$: $P + \frac{16}{P} = 10 \implies P^2 - 10P + 16 = 0 \implies (P-8)(P-2) = 0 \implies P = 8$ or $P = 2$.Case $1$: $\cos^2 x = \log_{16} 8 = \log_{2^4} 2^3 = \frac{3}{4} \implies \cos x = \pm \frac{\sqrt{3}}{2} \implies x = n\pi \pm \frac{\pi}{6}$.Case $2$: $\cos^2 x = \log_{16} 2 = \log_{2^4} 2^1 = \frac{1}{4} \implies \cos x = \pm \frac{1}{2} \implies x = n\pi \pm \frac{\pi}{3}$.Thus,$B = \{n\pi \pm \frac{\pi}{6}, n\pi \pm \frac{\pi}{3} \mid n \in Z\}$.$B - A$ removes the elements of $A$ from $B$,leaving $B - A = \{n\pi \pm \frac{\pi}{3} \mid n \in Z\}$.Expanding for $n=2k$ and $n=2k+1$: $x = 2k\pi \pm \frac{\pi}{3}$ and $x = (2k+1)\pi \pm \frac{\pi}{3} = 2k\pi + \pi \pm \frac{\pi}{3} = 2k\pi \pm \frac{2\pi}{3}$.Therefore,$B - A = \{x \in R \mid x = 2n\pi \pm \frac{\pi}{3}, 2n\pi \pm \frac{2\pi}{3}, n \in Z\}$.

If $A$ is the solution set of the equation $\cos ^2 x = \cos ^2 \frac{\pi}{6}$ and $B$ is the solution set of the equation $\cos ^2 x = \log _{16} P$ where $P + \frac{16}{P} = 10$,then $B - A =$

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