solve 2cos(2theta-pi)= -2 , -2 pi less or equal to theta less than 2 pi

Answer 1

2 cos(2Ø-π) = -2
cos (2Ø-π) = -1
2Ø-π = π
2Ø = 2π
Ø = π

the period of cos 2Ø = π , so adding/subtracting multiples of π will yield another answer

so solutions for -2π ≤ Ø ≤ 2π :
± π , ± 2π

Answer 2

Let x = theta
2 cos(2x - π) = -2

Divide everything by 2
cos(2x - π) = -1

Expand the left side by applying the difference of cosines:
cos(A ± B) = cos(A)cos(B) ∓ sin(A)sin(B)
Thus,
cos(2x)cos(π) + sin(2x)sin(π) = -1
cos(2x)*(-1) + sin(2x)*(0) = -1
-cos(2x) = -1
cos(2x) = 1
2x = cos^-1 (1)
x = cos^(1) / 2

Note that cos^-1 (1) = -2π or 0 or 2π
So, x = -π or 0 or π

Hope this helps~ `u`

Answer 3

Looks like both Jai and I both forgot an angle
I left out 0 and Jai left out 2π

Final answer:

for -2π ≤ Ø ≤ 2π

Ø = 0, ±π, ±2π