Sine and Cosine powered

$\begin{aligned} \sin^6 x + \cos^6 x & = \frac 14 & \small \blue{\text{Note that }a^3 + b^3 = (a+b)(a^2 - ab + b^2)} \\ (\sin^2 x + \cos^2 x) (\blue{\sin^4 x} - \sin^2 x \cos^2 x + \blue{\cos^4 x}) & = \frac 14 & \small \blue{\text{and }(a^2 + b^2 = (a+b)^2 - 2ab} \\ 1 \cdot (\blue{(\sin^2 x+\cos^2 x)^2 - 2\sin^2 x \cos^2 x} - \sin^2 x \cos^2 x) & = \frac 14 \\ 1- 3\sin^2 x \cos^2 x & = \frac 14 \\ 3\sin^2 x \cos^2 x & = \frac 34 \\ \frac 14 \sin^2 2x & = \frac 14 \\ \sin^2 2x & = 1 \\ \implies \sin 2x & = \pm 1 \\ 2x & = k\pi + \frac \pi 2 \\ \implies x & = \boxed{\frac {k\pi}2 + \frac \pi 4} \end{aligned}$

$(\sin^6 x+\cos^6x)=(\sin^2 x+\cos^2 x)^3 -3\sin^2 x \cos^2 x (\sin^2 x+\cos^2 x)$

$(\sin^2 x +\cos^2 x)^3 -3\sin^2 x \cos^2 x (\sin^2 x +\cos^2 x )$ = $\frac{1}{4}$

$1-3\sin^2 x \cos^2 x$ = $\frac{1}{4}$

$1-\frac{3}{4}\sin^2 (2x)$ = $\frac{1}{4}$

$\sin^2 (2x)$ = $1$

$\sin2x$ = $±1$

$2x$ = $kπ±(±\frac{π}{2})$

$x$ = $\boxed{k\frac{π}{2}+\frac{π}{4}}$

$k∈Z$ $:)$

We know that $s \sin(45°) = \cos(45°) = \frac{\sqrt{2}}{2}$ .

Then $\sin^6(45°) = \frac{1}{8}$

$\frac{1}{8} + \frac{1}{8} = \frac{1}{4}$

Now we just have just have to include all the $90°$ rotations over the axis

For this we use the expression $\frac{k\pi}{2}$ which will always be a multiple of $90°$

$\sin^6{x} + \cos^6{x} = \frac{1}{4}$ Using: $a^3 + b^3 = (a+b)(a^2+b^2-ab)$ : $(\sin^2{x} + \cos^2{x})(\sin^4{x} + \cos^4{x} - \sin^2{x} \cos^2{x}) = \frac{1}{4}$ Using: $\sin^2{x} + \cos^2{x}=1$ and adding and subtracting $2 \sin^2{x} \cos^2{x}$ in the LHS: $\sin^4{x} + \cos^4{x} +2 \sin^2{x} \cos^2{x} - 3 \sin^2{x} \cos^2{x}= \frac{1}{4}$ Using: $(a+b)^2 = a^2 + b^2 +2ab$ : $(\sin^2{x} + \cos^2{x})^2 - 3 \sin^2{x} \cos^2{x}= \frac{1}{4}$ $3 \sin^2{x} \cos^2{x}= 1 - \frac{1}{4}$ $\sin^2{x} \cos^2{x} = \frac{1}{4}$ $\sin^2{2x} = 1$ $\implies \sin{2x} = 1$ $\implies \sin{2x} = -1$

Therefore $2x$ must be any odd multiple of $\pi/2$ . In other words:

$2x = (2k+1) \frac{\pi}{2}$ $\boxed{x = \frac{k \pi}{2} + \frac{\pi}{4}}$

here, $k$ is an integer.