trigonometry math

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Determine the values of x, 0<x<2pi, if the identity is true: cos^6 x=1-sin^6 x

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The equation to be solved is (cos x)^6 = 1 - (sin x)^6

(cos x)^6 = 1 - (sin x)^6

=> (cos x)^6 + (sin x)^6 = 1

=> (cos x)^2^3 + (sin x)^2^3 = 1

use a^3 + b^2 = (a + b)(a^2 - ab + b^2)

=> [(cos x)^2 + (sin x)^2][(cos x)^4 - (cos x)^2* (sin x)^2 + (sin x)^4]

=> (cos x)^4 - (cos x)^2* (sin x)^2 + (sin x)^4

=> ((cos x)^2 + (sin x)^2)^2 - 2*(cos x)^2*(sin x)^2 - (cos x)^2* (sin x)^2

=> 1 - 3*(cos x)^2*(sin x)^2 = 1

=> - 3*(cos x)^2*(sin x)^2 = 0

=> (cos x)^2 (1 - (cos x)^2) = 0

=> cos x = 0 and cos x = 1 and cos x = -1

x = arc cos 0 = pi/2 and 3*pi/2

x = arc cos 1 = 0, which can be eliminated

x = arc cos -1 = pi

The solution of the equation is (pi/2, pi, 3*pi/2)

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