1. We need to solve $4\cos\theta-3\sin\theta=1, \qquad \theta\in\left[0,\frac{\pi}{4}\right].$ Let $c=\cos\theta,\qquad s=\sin\theta.$ Then $4c-3s=1$ and also $c^2+s^2=1.$

2. Express $s$ in terms of $c$: $s=\frac{4c-1}{3}.$ Substitute into $c^2+s^2=1$: $c^2+\left(\frac{4c-1}{3}\right)^2=1.$

3. Simplify: $c^2+\frac{16c^2-8c+1}{9}=1$ $9c^2+16c^2-8c+1=9$ $25c^2-8c-8=0.$

4. Solve the quadratic: $c=\frac{8\pm\sqrt{64+800}}{50}$ $=\frac{8\pm\sqrt{864}}{50}$ $=\frac{8\pm 12\sqrt6}{50}$ $=\frac{4\pm 6\sqrt6}{25}.$

5. Since $\theta\in\left[0,\frac{\pi}{4}\right]$, we have $\cos\theta\in\left[\frac{1}{\sqrt2},1\right],$ so $\cos\theta$ must be positive and fairly large. Thus $\cos\theta=\frac{4+6\sqrt6}{25}.$ (The other root is negative, so it is impossible.)

6. Now compare with the options. Rationalize option D:

This is exactly $\frac{4+6\sqrt6}{25}.$

Hence, $\boxed{\cos\theta=\frac{4}{3\sqrt6-2}}.$ So the correct option is D.