#1**+1 **

Use the sum of cubes factorization to get:

Suppose that sin x = x and cos x = y.

x^6 + y^6 = (x^2 + y^2)((x^2 + y^2)^2 - 3x^2y^2)

x^2 + y^2 = 1 as sin^2 x + cos^2 x is always one by the pythagorean identity.

By sin(2x) = 2/3, we can find that 2sin(x)cos(x) = 2/3

Leaving us with xy = 1/3 which finally gives us x^2y^2 = 1/9

Substituing all values, we result in:

(1)(2/3) = \(\boxed{\frac23}\)

Voldemort Aug 5, 2022

#1**+1 **

Best Answer

Use the sum of cubes factorization to get:

Suppose that sin x = x and cos x = y.

x^6 + y^6 = (x^2 + y^2)((x^2 + y^2)^2 - 3x^2y^2)

x^2 + y^2 = 1 as sin^2 x + cos^2 x is always one by the pythagorean identity.

By sin(2x) = 2/3, we can find that 2sin(x)cos(x) = 2/3

Leaving us with xy = 1/3 which finally gives us x^2y^2 = 1/9

Substituing all values, we result in:

(1)(2/3) = \(\boxed{\frac23}\)

Voldemort Aug 5, 2022