Answer:
[tex]\displaystyle x = \frac{\pi}{4} + k\, \pi[/tex] for integer [tex]k[/tex] (including negative numbers.)
Step-by-step explanation:
Pythagorean Identity: [tex]\sin^{2}(x) + \cos^{2}(x) = 1[/tex]. Equivalently, [tex]\cos^{2}(x) = 1 - \sin^{2}(x)[/tex].
Rewrite the original equation and apply this substitution to eliminate [tex]\cos(x)[/tex]:
[tex]\displaystyle \sin^{6}(x) + \cos^{6}(x) = \frac{1}{4}[/tex].
[tex]\displaystyle (\sin^{2}(x))^{3} + (\cos^{2}(x))^{3} = \frac{1}{4}[/tex].
[tex]\displaystyle (\sin^{2}(x))^{3} + (1 - \sin^{2}(x))^{3} = \frac{1}{4}[/tex].
Let [tex]y = \sin(x)[/tex] ([tex]-1 \le y \le 1[/tex].) The original equation is equivalent to the following equation about [tex]y[/tex]:
[tex]\displaystyle y^{6} + (1 - y^{2})^{3} = \frac{1}{4}[/tex].
Expand the cubic binomial in the equation:
[tex]\displaystyle y^{6} + 1 - 3\, y^{2} + 3\, (y^{2})^{2} - (y^{2})^{3} = \frac{1}{4}[/tex].
[tex]\displaystyle y^{6} + 1 - 3\, y^{2} + 3\, y^{4} - y^{6} = \frac{1}{4}[/tex].
Simplify to obtain:
[tex]\displaystyle 1 - 3\, y^{2} + 3\, y^{4} = \frac{1}{4}[/tex].
Rearrange and simplify:
[tex]12\, y^{4} - 12\, y^{2} + 3 = 0[/tex].
[tex]3\, (2\, y^{2} - 1)^{2} = 0[/tex].
[tex]2\, y^{2} - 1 = 0[/tex].
[tex]\displaystyle y^{2} - \frac{1}{2} = 0[/tex].
Solve for [tex]y[/tex]:
Either [tex]\displaystyle y = \frac{1}{\sqrt{2}}[/tex] or [tex]\displaystyle y = -\frac{1}{\sqrt{2}}[/tex].
If [tex]\displaystyle \sin(x) = y = \frac{1}{\sqrt{2}}[/tex], then [tex]\displaystyle x = \frac{\pi}{4} + 2\, k\,\pi[/tex] for all [tex]k\in \mathbb{Z}[/tex].
On the other hand, if [tex]\displaystyle \sin(x) = y = \frac{1}{\sqrt{2}}[/tex], then [tex]\displaystyle x = \frac{3\, \pi}{4} + 2\, k\,\pi = \frac{\pi}{4} + (2\, k + 1) \, \pi[/tex] for all [tex]k\in \mathbb{Z}[/tex].
Combine both situations to obtain:
[tex]\displaystyle x = \frac{\pi}{4} + 2\, k\, \pi[/tex] for all [tex]k \in \mathbb{Z}[/tex].
