Trigonometric equation result differs from given i've got an equation:
$$\sin^6(x) + \cos^6(x) = 0.25$$
and i'm trying to solve it using the sum of cubes formula, like this:
$$ (\sin^2(x))^3 + (\cos^2(x))^3 = 0.25 $$
$$ (\sin^2(x) + \cos^2(x))^2 (\sin^4(x) - \sin^2(x)\cos^2(x) + \cos^4(x)) = 0.25 $$
$$ 1 - 3\sin^2(x)\cos^2(x) = 0.25 $$
$$ -3 \sin^2(x)\cos^2(x) = -\frac{3}{4} $$
$$ \sin^2(2x) = 1 $$
$$ \sin^2(2x) = \sin^2(2x) + \cos^2(2x) $$
$$ \cos^2(2x) = 0 $$
and here it must be $x = \frac{\pi n}{2} \pm \frac{\pi}{4} $
but solution is $ x = \pi n \pm \frac{\pi}{4}$
What's wrong?

Both the answers are correct. Since, both represent the same solution set. 

This is so because each odd number can be either represented as $2n+1$ and $2m-1$, or represented as $4p+1$ or $4q-1$. And hence $$2x=(2n\pm 1)\frac{\pi}{2} \implies \frac{n\pi}{2} \pm \frac{\pi}{4} $$
$$ \text{also} \; 2x =(4n\pm 1)\frac{\pi}{2} \implies n\pi \pm \frac{\pi}{4} $$ So, both the solution set are basically the same.
It although doesn't sound to be convincing, so you can try to write a few elements of your set 
$$\mathrm S_1= \frac{n \pi}{2} \pm \frac{\pi}{4} =\left \{\ldots -\frac{3\pi}{4}, -\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4} \ldots\right \} $$
$$\mathrm S_2= n \pi\pm \frac{\pi}{4} =\left \{\ldots -\frac{3\pi}{4}, -\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4} \ldots\right \} $$
