The problem is quite simple, I just need to make sure there is no quick and obvious way I am missing.
Find the minimum positive root of the equation $\sin{3x}=a$ ($a>0$) given that there exist two roots such that their difference equals $\frac{\pi}{4}$.
How I solved the problem: We know that the equation has roots: $x_1$ and $\frac{\pi}{4}+x_1$. Plugging that into the original equation we get: $$\sin\left(\frac{3\pi}{4} + 3x_1\right) = a$$ from which we can find the positive value of $a$ that equals $\frac{\sqrt{2-\sqrt{2}}}{2}$. Having the value of $a$, we can now plug it into $\sin{3x}=a$ and find the desired minimum positive root.
Is there a better way (without graphs) to solve this problem?
The smaller positive roots occur in the interval $\left[0,\frac{1}{3}\pi\right]$ when $a>0$. If the smallest is $x_1$, then because of symmetry, the next smallest will be $\frac{\pi}{3}-x_1$.
If the two roots are to be a distance apart of $\frac{\pi}{4}$ then
$$ \left(\frac{\pi}{3}-x_1\right)-x_1=\frac{\pi}{4} $$
So $x_1=\frac{\pi}{24}$.
ADDENDUM: If you do not wish to argue from the symmetry of the graph about $x=\frac{\pi}{6}$ you may instead note that
$$\sin\left(3\left(\frac{\pi}{3}-x_1\right)\right)=\sin(3x_1)$$