Determine the values of x, 0

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}.