To solve sin^6x+cos^6x = 1.

We know a^3+b^3 = (a+b)(a^2+b^2-ab)

Therefore,

LHS= sin^6x+cos^6x = (sin^2x+cos^2){sin^4x+cos^4x+sin^2-cos^2x) .

sin^6x +cos^6x = (sin^4x+cos^4x -sin^2cos^2x), as sin^2x+cos^2x = 1.

sin^4x+cos^4x = (sin^2+cos^2x)^2 -2sin^2cos^2x = 1- 2in^2xcos^2x.

Therefore sin^6x +cos^6x = {1-2sin^2xcos^2x -sin^2xcos^2x.}= 1-3sin^2xcos^2x.

Therefore sin^6x +cos^6x = 1 implies 1-3sin^2xcos^2x = 1.

Or -3sin^2xcos^2x = 0

Sinc2 -3 cannot be zero, sin^2 x = 0 cos^2 = 0.

If sin^2 = zero, the x = npi.

If cos^2 = 0, then x = 2npi -pi/3 or x = 2npi +oi/2.

We'll try to use the fundamental formula of trigonometry, to solve the equation:

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

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 put each factor as zero:

(sin x)^2 = 0

sin x = 0

x = arcsin 0 + k*pi

x = 0

x = pi

x = 2pi

(cos x)^2 = 0

cos x = 0

x = pi/2

x = 3pi/2

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