Step 1: Simplify the inner cosine inverse.

We know that \(\cos^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{6}\).

Step 2: Simplify the tangent function.

So, \(2\cos^{-1}\frac{\sqrt{3}}{2} = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3}\). Then, \(\tan(2\cos^{-1}\frac{\sqrt{3}}{2}) = \tan(\frac{\pi}{3}) = \sqrt{3}\).

Step 3: Substitute into the original equation.

The equation becomes \(\sin^{-1}[k\sqrt{3}] = \frac{\pi}{3}\).

Step 4: Solve for k.

Taking sine of both sides, we get \(k\sqrt{3} = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\). Therefore, \(k = \frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}\).