If $f: R \rightarrow[-1,1]$ and $g: R \rightarrow A$ are two surjective mappings and $\sin \left(g(x)-\frac{\pi}{3}\right)=\frac{f(x)}{2} \sqrt{4-f^2(x)}$, then $A=$

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Let $f(x)=y$, then $ \begin{aligned} & \sin \left(g(x)-\frac{\pi}{3}ight)=\frac{f(x)}{2} \sqrt{4-f^2(x)}=\frac{y}{2} \sqrt{4-y^2}=t 	ext { (let) } \ & \Rightarrow \quad y^2-\frac{y^4}{4}=t^2 \quad \Rightarrow \quad 4 y^2-y^4=4 t^2 \ & \Rightarrow \quad\left(y^2-2^2=-4 t^2+4 \quad \Rightarrow \quad t^2=1-\frac{1}{4}\left(y^2-2^2ight)^2ight. \ & \because \quad f(x)=y \in[-1,1] \quad \Rightarrow \quad y^2 \in[0,1] \ & 	herefore \quad t^2 \in\left[0, \frac{3}{4}ight] \quad \Rightarrow \quad t \in\left[-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}ight] \ & 	ext { So, } \sin \left(g(x)-\frac{\pi}{3}ight) \in\left[-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}ight] \ & \Rightarrow \quad g(x)-\frac{\pi}{3} \in\left[-\frac{\pi}{3}, \frac{\pi}{3}ight] \Rightarrow g(x) \in\left[0, \frac{2 \pi}{3}ight] \end{aligned} $

If $f: R \rightarrow[-1,1]$ and $g: R \rightarrow A$ are two surjective mappings and $\sin…

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