# 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)**

We'll recall the Pythagorean identity:

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

We'll apply the formula of binomial raised to cube:

(a+b)^3 = a^3 + b^3 + 3a*b*(a+b)

a = (sin x)^2 and b = (cos x)^2

We'll write (sin x)^6 + (cos x)^6 = [(sin x)^2 + (cos x)^2]^3 - 3(sin x)^2*(cos x)^2)[(sin x)^2 + (cos x)^2]

We'll substitute the sum (sin x)^2 + (cos x)^2 by 1 and we'll get:

(sin x)^6 + (cos x)^6 = 1^3 - 3*1*[(sin x)^2*(cos x)^2]

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

We'll re-write the equation:

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

We'll eliminate like terms and we'll get:

3*[(sin x)^2*(cos x)^2] = 0

We'll divide by 3:

[(sin x)^2*(cos x)^2] = 0

We'll cancel each factor, one by one:

(sin x)^2 = 0

sin x = 0

x = arcsin 0 + k*pi

x = 0 , but x > 0, so we'll reject the solution

x = pi

x = 2pi , but x < 2pi, so we'll reject the solution

(cos x)^2 = 0

cos x = 0

x = pi/2

x = 3pi/2

**The possible solutions of the equation, for 0<x<2pi, are: { pi/2 , pi, 3pi/2}.**