Trigonometric Equations

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=‌

1

4

, which are:
B−A={x∈ℝlvert‌x=2nπ±‌

π

3

.,2nπ±‌

2π

3

,n∈ℤ}