Question

Sin^6x + cos^6x = 1/4

Answer 1

`\displaystyle x = (\pi)/(4) + k\, \pi` for integer
`k` (including negative numbers.)

Explanation:

Pythagorean Identity:
`\sin^(2)(x) + \cos^(2)(x) = 1`. Equivalently,
`\cos^(2)(x) = 1 - \sin^(2)(x)`.

Rewrite the original equation and apply this substitution to eliminate
`\cos(x)`:

`\displaystyle \sin^(6)(x) + \cos^(6)(x) = (1)/(4)`.

`\displaystyle (\sin^(2)(x))^(3) + (\cos^(2)(x))^(3) = (1)/(4)`.

`\displaystyle (\sin^(2)(x))^(3) + (1 - \sin^(2)(x))^(3) = (1)/(4)`.

Let
`y = \sin(x)` (
`-1 \le y \le 1`.) The original equation is equivalent to the following equation about
`y`:

`\displaystyle y^(6) + (1 - y^(2))^(3) = (1)/(4)`.

Expand the cubic binomial in the equation:

`\displaystyle y^(6) + 1 - 3\, y^(2) + 3\, (y^(2))^(2) - (y^(2))^(3) = (1)/(4)`.

`\displaystyle y^(6) + 1 - 3\, y^(2) + 3\, y^(4) - y^(6) = (1)/(4)`.

Simplify to obtain:

`\displaystyle 1 - 3\, y^(2) + 3\, y^(4) = (1)/(4)`.

Rearrange and simplify:

`12\, y^(4) - 12\, y^(2) + 3 = 0`.

`3\, (2\, y^(2) - 1)^(2) = 0`.

`2\, y^(2) - 1 = 0`.

`\displaystyle y^(2) - (1)/(2) = 0`.

Solve for
`y`:

Either
`\displaystyle y = (1)/(√(2))` or
`\displaystyle y = -(1)/(√(2))`.

If
`\displaystyle \sin(x) = y = (1)/(√(2))`, then
`\displaystyle x = (\pi)/(4) + 2\, k\,\pi` for all
`k\in \mathbb{Z}`.

On the other hand, if
`\displaystyle \sin(x) = y = (1)/(√(2))`, then
`\displaystyle x = (3\, \pi)/(4) + 2\, k\,\pi = (\pi)/(4) + (2\, k + 1) \, \pi` for all
`k\in \mathbb{Z}`.

Combine both situations to obtain:

`\displaystyle x = (\pi)/(4) + 2\, k\, \pi` for all
`k \in \mathbb{Z}`.