(1/2)cos(4x-pi), -pi<=x<=pi what are the turning points.?

-π ≤ x ≤ π
-4π ≤ 4x ≤ 4π
-5π ≤ (4x - π) ≤ 3π

y = (1/2) cos(4x - π)
y' = (1/2) (4) sin(4x - π) = 2 sin(4x - π)
y" = 2 (4) cos(4x - π) = 8 cos(4x - π)

When y' = 0 :
2 sin(4x - π) = 0
sin(4x - π) = 0
4x - π = -5π, -4π, -3π, -2π, -π, 0, π, π, 3π
4x = -4π, -3π, -2π, -π, 0, π, 2π, 3π, 4π
x = -π, -3π/4, -π/2, -π/4, 0, π/4, π/2, 3π/4, π

When x = -π, -π/2, 0, π/2, π :
y' = 0
y" = 8 cos(-4x - π) = -8 < 0
Maximum points at x = -π, -π/2, 0, π/2, π

When x = -3π/4, -π/4, π/4, 3π/4 :
y = 0
y" = 8 cos[-4x - π] = 8 > 0
Minimum points at x = -3π/4, -π/4, π/4, 3π/4