Trigonometric equation result differs from given

Question

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?

Answer 1

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

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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

$$\mathrm S_1= \frac{n \pi}{2} \pm \frac{\pi}{4} =\left \{\ldots -\frac{3\pi}{4}, -\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4} \ldots\right \} $$

$$\mathrm S_2= n \pi\pm \frac{\pi}{4} =\left \{\ldots -\frac{3\pi}{4}, -\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4} \ldots\right \} $$

Answer 2

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