To solve this problem, let's consider the solution sets of the given equations step by step. Set A: We start with the equation: cos2x=cos2
π
6
. Since cos
π
6
=
√3
2
, we have cos2
π
6
=
3
4
. The solutions to this equation are: x=nπ±
π
6
, where n∈ℤ Set B: We consider the equation: cos2x=log16P. Given P+
16
P
=10, finding P involves solving the quadratic equation p2−10p+16=0. Factoring gives: (p−8)(p−2)=0, so p=8 or p=2. Thus, cos2x=log168 or cos2x=log162. For p=8,log168=
3
4
(. since
3
4
=cos2(
π
6
)). For p=2,log162=
1
4
. Hence, cos2x=
3
4
aligns with cos2x=cos2
π
6
. For cos2x=
1
4
, the solutions are: x=2nπ±
π
3
and x=2nπ±
2π
3
, where n∈ℤ Finding B−A : Set B includes solutions for both cos2x=
1
4
and cos2x=
3
4
. Set A only includes solutions for cos2x=
3
4
. Therefore, the difference B−A consists of solutions only when cos2x=