Suppose $$\theta \in [0, \frac{\pi}{4}]$$ is a solution of $$4\cos\theta - 3\sin\theta = 1$$. Then $$\cos\theta$$ is equal to :

Rearrange and Square: $$4\cos\theta - 1 = 3\sin\theta$$

$$(4\cos\theta - 1)^2 = (3\sin\theta)^2$$

$$16\cos^2\theta - 8\cos\theta + 1 = 9(1 - \cos^2\theta)$$

$$16\cos^2\theta - 8\cos\theta + 1 = 9 - 9\cos^2\theta$$

$$25\cos^2\theta - 8\cos\theta - 8 = 0$$

Using Quadratic Formula

$$\cos\theta = \frac{8 \pm \sqrt{64 - 4(25)(-8)}}{2(25)} = \frac{8 \pm \sqrt{864}}{50} = \frac{8 \pm 12\sqrt{6}}{50} = \frac{4 \pm 6\sqrt{6}}{25}$$

Option C is $$\frac{4}{3\sqrt{6}-2}$$. Multiplying by the conjugate $$\frac{3\sqrt{6}+2}{3\sqrt{6}+2}$$:

$$\frac{4(3\sqrt{6}+2)}{54-4} = \frac{4(3\sqrt{6}+2)}{50} = \frac{6\sqrt{6}+4}{25}$$

This matches our positive root.