If $$3 \sec \theta + 4 \cos \theta - 4\sqrt{3} = 0$$ where $$\theta$$ is an acute angle then the value of $$\theta$$ is:

$$3\sec\theta+4\cos\theta-4\sqrt{3}=0$$

$$\frac{3}{\cos\theta\ }+4\cos\theta-4\sqrt{3}=0$$

$$4\cos^2\theta-4\sqrt{3}\cos\theta\ +3=0$$

$$4\cos^2\theta-2\sqrt{3}\cos\theta\ -2\sqrt{3}\cos\theta+3=0$$

$$2\cos\theta\ \left(2\cos\theta-\sqrt{3}\right)\ -\sqrt{3}\left(2\cos\theta-\sqrt{3}\right)=0$$

$$\ \left(2\cos\theta-\sqrt{3}\right)\left(2\cos\theta-\sqrt{3}\right)=0$$

$$\ \left(2\cos\theta-\sqrt{3}\right)^2=0$$

$$\ 2\cos\theta-\sqrt{3}=0$$

$$\cos\theta=\frac{\sqrt{3}}{2}$$

$$\theta=30^{\circ\ }$$

Hence, the correct answer is Option B