Because both $\sin(a)$ and $\cos(a)$ are non-zero on the given interval, we may multiply through by $\sin(a)\cos(a)$ to obtain:
$$\left(\sqrt{3}-1\right)\cos(a)+\left(\sqrt{3}+1\right)\sin(a)=4\sqrt{2}\sin(a)\cos(a)$$
Using a linear combination on the left and the double-angle identity for sine on the right, there results:
$$2\sqrt{2}\sin\left(a+\frac{\pi}{12}\right)=2\sqrt{2}\sin(2a)$$
or:
$$\sin\left(a+\frac{\pi}{12}\right)=\sin(2a)$$
In light of the identity $\sin(\pi-x)=\sin(x)$, we have 2 cases to consider:
(i) $$\sin\left(a+\frac{\pi}{12}\right)=\sin(2a)$$
From this, we obtain:
$$a+\frac{\pi}{12}=2a$$
$$a=\frac{\pi}{12}$$
(ii) $$\sin\left(\pi-\left(a+\frac{\pi}{12}\right)\right)=\sin(2a)$$
From this we obtain:
$$\pi-\left(a+\frac{\pi}{12}\right)=2a$$
$$3a=\frac{11\pi}{12}$$
$$a=\frac{11\pi}{36}$$
Now, we must also consider (because of the periodicity of the sine function):
(iii) $$\sin\left(\pi-\left(a+\frac{\pi}{12}\right)\right)=\sin(2a-2\pi)$$
From this we obtain:
$$\pi-\left(a+\frac{\pi}{12}\right)=2a-2\pi$$
$$3a=\frac{35\pi}{12}$$
$$a=\frac{35\pi}{36}$$
This solution is outside of the given interval.
In light of the identities $\sin(x+\pi)=-\sin(x)$ and $\sin(-x)=-\sin(x)$, we have 1 more case to consider:
(iv) $$\sin\left(a+\frac{\pi}{12}+\pi\right)=\sin(-2a)$$
From this, we obtain:
$$a+\frac{\pi}{12}+\pi=-2a$$
$$3a=-\frac{13\pi}{12}$$
$$a=-\frac{13\pi}{36}$$
This solution is outside of the given interval.
All other solutions resulting from periodicity are outside the given interval.
Thus, the only two solutions in the given interval are:
$$a\in\left\{\frac{\pi}{12},\frac{11\pi}{36}\right\}$$