Find all x, -4pi < x < 6pi, such that [cos(x/3)]^4 + [sin(x/3)]^4 = 1

PLEASEEE HELPP! :(

realizing that

m^4 + n^4 = (m^2 + n^2)^2 - 2(m^2)(n^2)

we can rewrite your equation as

((cos x/3)^2 + (sin x/3)^2)^2 - 2(sin x/3)^2 (cos x/3)^2 = 1

(1)^2 - 2(sin x/3)^2 (cos x/3)^2 = 1

(sin x/3)(cos x/3)=0
so x/3 = 0,pi,2pi or x = pi/2, 3pi/2

those are the answers from 0 to 2pi

now fill in the rest from -4pi to 6pi