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?

Observe that $$\dfrac{\pi n}2\pm\dfrac\pi4=\dfrac{\pi(2n\pm1)}4$$
Now set a few values of $n$ to check overlap which is evident as $2n+1=2(n+1)-1$
Actually, $\cos y=0\implies y=(2m+1)\dfrac\pi2$ where $m$ is any integer
