You barely need trigonometry!

$\begin{aligned} (\sin^2 x + \cos^2 x)^3 & = \sin^6 x + 3\sin^4 x \cos^2 x + 3 \sin^2 x \cos^4 x + \cos^6 x \\ 1 & = \sin^6 x + \cos^6 x + 3 \sin^2 x \cos^2 x(\sin^2 x + \cos^2 x) \\ 1 & = \sin^6 x + \cos^6 x + \frac 34 \sin^2 2x \end{aligned}$

$\begin{aligned} \implies \sin^6 x + \cos^6 x & = 1 - \frac 34 \sin^2 2x \\ 1 - \frac 34 \sin^2 2x & = \frac 14 \\ \sin^2 2x & = 1 \\ \sin 2x & = 1 & \small \color{#3D99F6} \text{for } 0^\circ < x < 90^\circ \\ 2x & = 90^\circ \\ \implies x & = \boxed{45}^\circ \end{aligned}$

(sinx)^6+(cox)^6=[(sinx) ^2+(cox)^2]^3 - 3(sinx)^2(cosx)^2[(sinx)^2+(cosx)^2]=1-3(sinx)^2(cosx)^2. Therefore 1-3(sinx)^2(cosx)^2=(1/4). Or 4(sinx)^2(cosx)^2=1 or sin(2x)=1 (since x<90 degrees, therefore sin(2x)>0) or 2x=90 degrees or x=45 degrees .

With the following two conditions known, it's easy to solve this problem:

1. Assume $sin^{2}x$ = $a$ , and

2. $sin^{2}x + cos^{2}x$ = $1$

Substituting (1) into the equation gives $a^{3}$ + $(1-a)^{3}$ = $\frac{1}{4}$

By expansion, $a^{3} + (1-3a+3a^{2}-a^{3})$ = $\frac{1}{4}$

$3a^{2}-3a+\frac{3}{4}$ = $0$

$a^{2}-a+\frac{1}{4} = 0$

$(a^-\frac{1}{2})^{2} = 0$

$a = \frac{1}{2}$

Since $sin^{2}x = \frac{1}{2}$ , $sinx = \frac{\sqrt2}{2}$

$x = \boxed{45^{\circ}}$

Here is a lazy solution: the question only states 1 x value with the interval that makes the equation true. The sin and cos graphs for the given interval are positive and symmetrical about x = 45°,the same goes for their 6 powers. Now if there is a solution other than 45°, there will be 2 solutions one x, and the other 90° - x. But since the question only asks for one solution, x must then be 45°. Substituting this in show sthat it is a solution