solve cos(2theta-pi)=squareroot2/2?

For 0 ≤ θ < 2π
Then 0 ≤ 2θ < 4π
and thus -π ≤ (2θ - π) < 3π

cos(2θ - π) = (√2)/2
cos(2θ - π) = cos(π/4)
2θ - π = -π/4, π/4, 2π-(π/4), 2π+(π/4)
2θ = 3π/4, 5π/4, 11π/4, 13π/4
θ = 3π/8, 5π/8, 11π/8, 13π/8

- cos 2Ө = - √2 / 2 = - 1 / √2
cos 2Ө = 1 / √2
2Ө = π/4 . 7π/4 . 9π/4 . 15π/4
Ө = π/8 . 7π/8 . 9π/8 . 15π/8

cos(2x-pi) = sqrt(2)/2
cos(-(pi-2x)) = sqrt(2)/2
cos(pi-2x) = sqrt(2)/2 ............. since cos(-u) = cos(u)
-cos(2x) = sqrt(2)/2 ............ since cos(pi-u) = -cos(u)
cos(2x) = -sqrt(2)/2

then 2x = 3pi/4 , 5pi/4
x = 3pi/8, 5pi/8

2*theta - pi = pi/4, 7*pi/4, -pi/4, -7pi/4, 9pi/4, - 9pi/4, 15pi/4, -15pi/4, etc.

theta = 5*pi/8, 11*pi/8, 13*pi/8; these are the only solutions on the interval [0,2*pi] but I don't know what interval they gave you.