Solve 4 sin² x+4√(2) cos x -6 = 0 for all real values of x

a. (3π/4) + (2πk), (5π/4) + (2πk)
b. (π/4) + (2πk), (7π/4) + (2πk)
c. (π/4) + (2πk), (5π/4) + (2πk)
d. (3π/4) + (2πk), (7π/4) + (2πk)

sin(x)^2 = 1 - cos(x)^2

4sin(x)^2 + 4 * sqrt(2) * cos(x) - 6 =>
2 * sin(x)^2 + 2 * sqrt(2) * cos(x) - 3 =>
2 * (1 - cos(x)^2) + 2 * sqrt(2) * cos(x) - 3 =>
2 - 2cos(x)^2 + 2 * sqrt(2) * cos(x) - 3 =>
-2cos(x)^2 + 2 * sqrt(2) * cos(x) - 1

-2cos(x)^2 + 2 * sqrt(2) * cos(x) - 1 = 0
2cos(x)^2 - 2sqrt(2) * cos(x) + 1 = 0

cos(x) = (2 * sqrt(2) +/- sqrt(8 - 8)) / 4
cos(x) = (2 * sqrt(2)) / 4
cos(x) = sqrt(2)/2
x = (pi/4) + 2pi * k , (7pi/4) + 2pi * k