The number of roots of the equation, $$(81)^{\sin^2 x} + (81)^{\cos^2 x} = 30$$ in the interval $$[0, \pi]$$ is equal to:

We need to solve $$(81)^{\sin^2 x} + (81)^{\cos^2 x} = 30$$ in the interval $$[0, \pi]$$.

Let $$t = (81)^{\sin^2 x} = 3^{4\sin^2 x}$$. Since $$\cos^2 x = 1 - \sin^2 x$$, we have $$(81)^{\cos^2 x} = 81/t$$. The equation becomes $$t + \frac{81}{t} = 30$$, which simplifies to $$t^2 - 30t + 81 = 0$$.

Solving: $$t = \frac{30 \pm \sqrt{900 - 324}}{2} = \frac{30 \pm \sqrt{576}}{2} = \frac{30 \pm 24}{2}$$. So $$t = 27$$ or $$t = 3$$.

Case 1: $$3^{4\sin^2 x} = 3 = 3^1$$, giving $$\sin^2 x = \frac{1}{4}$$, so $$\sin x = \frac{1}{2}$$ (since $$\sin x \geq 0$$ in $$[0, \pi]$$). This gives $$x = \frac{\pi}{6}$$ and $$x = \frac{5\pi}{6}$$.

Case 2: $$3^{4\sin^2 x} = 27 = 3^3$$, giving $$\sin^2 x = \frac{3}{4}$$, so $$\sin x = \frac{\sqrt{3}}{2}$$. This gives $$x = \frac{\pi}{3}$$ and $$x = \frac{2\pi}{3}$$.

Therefore, there are 4 roots in the interval $$[0, \pi]$$.