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We are given the equation $sin^{-1}[k~tan(2~cos^{-1}\frac{\sqrt{3}}{2})]=\frac{\pi}{3}$. First, we evaluate $cos^{-1}\frac{\sqrt{3}}{2}$. Since $cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, we have $cos^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{6}$.
Now we substitute this value into the equation: $sin^{-1}[k~tan(2 \cdot \frac{\pi}{6})] = \frac{\pi}{3}$ $sin^{-1}[k~tan(\frac{\pi}{3})] = \frac{\pi}{3}$ Since $tan(\frac{\pi}{3}) = \sqrt{3}$, we have $sin^{-1}[k\sqrt{3}] = \frac{\pi}{3}$
Taking the sine of both sides, we get $sin(sin^{-1}[k\sqrt{3}]) = sin(\frac{\pi}{3})$ $k\sqrt{3} = \frac{\sqrt{3}}{2}$ Dividing both sides by $\sqrt{3}$, we get $k = \frac{1}{2}$
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