If $$\int_0^{\pi/3} \cos^4 x \, dx = a\pi + b\sqrt{3}$$, where $$a$$ and $$b$$ are rational numbers, then $$9a + 8b$$ is equal to:

We need to evaluate $$\int_0^{\pi/3} \cos^4 x \, dx$$ and express it as $$a\pi + b\sqrt{3}$$, then find $$9a + 8b$$.

Since $$\cos^4 x = (\cos^2 x)^2$$, applying the power-reduction formula gives

$$\cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1 + 2\cos 2x + \cos^2 2x}{4}.$$

Now using $$\cos^2 2x = \frac{1 + \cos 4x}{2}$$ and substituting yields

$$\cos^4 x = \frac{1 + 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{3 + 4\cos 2x + \cos 4x}{8}.$$

Next, integrate term by term:

$$\int_0^{\pi/3} \cos^4 x \, dx = \frac{1}{8}\int_0^{\pi/3}(3 + 4\cos 2x + \cos 4x)\,dx$$

$$= \frac{1}{8}\Bigl[3x + 2\sin 2x + \frac{\sin 4x}{4}\Bigr]_0^{\pi/3}.$$

At $$x = \pi/3$$, we have

$$3 \cdot \frac{\pi}{3} = \pi,\quad 2\sin\frac{2\pi}{3} = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3},\quad \frac{1}{4}\sin\frac{4\pi}{3} = \frac{1}{4}\bigl(-\frac{\sqrt{3}}{2}\bigr) = -\frac{\sqrt{3}}{8}.$$

At $$x = 0$$ all terms vanish.

Substituting these values gives

$$\frac{1}{8}\Bigl(\pi + \sqrt{3} - \frac{\sqrt{3}}{8}\Bigr) = \frac{1}{8}\Bigl(\pi + \frac{7\sqrt{3}}{8}\Bigr) = \frac{\pi}{8} + \frac{7\sqrt{3}}{64}.$$

Therefore $$a = \frac{1}{8}$$ and $$b = \frac{7}{64}$$.

Finally, computing $$9a + 8b$$ gives

$$9a + 8b = 9 \cdot \frac{1}{8} + 8 \cdot \frac{7}{64} = \frac{9}{8} + \frac{56}{64} = \frac{9}{8} + \frac{7}{8} = 2.$$

The correct answer is Option 1: 2.