262hint : use AM-GM inequality
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\hspace{-16}\bf{\mathbb{F}}$ind Real values of $\bf{x}$ that satisfy the equation\\\\ $\bf{4^{x.\sin^{-1}(x)}+4^{x.\cos^{-1}(x)}=2^{\frac{2+\pi.x}{2}}}$
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1708\hspace{-16}$Yes $\bf{\mathbb{A}}$ditiya You are Right.\\\\ Yes $\bf{\mathbb{R}}$ahul Here $\bf{x\in \left[-1,1\right]}$\\\\ I have Solved Like This way.....\\\\ Here $\bf{4^{x.\sin^{-1}(x)}+4^{x.\cos^{-1}(x)}=2^{\frac{2+\pi.x}{2}}}$\\\\ Using $\bf{\cos^{-1}(x)=\frac{\pi}{2}-\sin^{-1}(x)}\;\forall x\in [-1,1]$ \\\\ Let $\bf{4^{x\sin^{-1}(x)}=\mathbb{A}}$ and $\bf{2^{\frac{\pi.x}{2}}=\mathbb{B}}$\\\\ So $\bf{\mathbb{A}+\frac{\mathbb{B}^2}{\mathbb{A}}=2\mathbb{B}\Leftrightarrow \left(\mathbb{A}-\mathbb{B}\right)^2=0}$\\\\ So $\bf{\mathbb{A=B}\Leftrightarrow 2^{2x.\sin^{-1}(x)}=2^{\frac{\pi.x}{2}}}$\\\\ So $\bf{2x.\sin^{-1}(x)=\frac{\pi.x}{2}=0}$\\\\ So $\boxed{\boxed{\bf{x=0\;\;,\frac{1}{\sqrt{2}}}}}$
1708\hspace{-16}$Using Aditiya Hint......\\\\ Using $\bf{A.M\geq G.M}$\\\\\\ $\bf{4^{x.\sin^{-1}(x)}\;,4^{x.\cos^{-1}(x)}>0\;\forall x\in [-1,1]}$\\\\\\ So $\bf{\frac{4^{x.\sin^{-1}(x)}+4^{x.\cos^{-1}(x)}}{2}\geq \sqrt{4^{x.\sin^{-1}(x)}.4^{x.\sin^{-1}(x)}}}$\\\\\\ $\bf{\frac{4^{x.\sin^{-1}(x)}+4^{x.\cos^{-1}(x)}}{2}\geq \sqrt{4^{x.\left\{\sin^{-1}(x)+{\cos^{-1}(x)}\right\}}}}$\\\\\\ Using $\bf{\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}}$\\\\\\ So $\bf{4^{x.\sin^{-1}(x)}+4^{x.\cos^{-1}(x)}\geq 2^{\frac{2+\pi.x}{2}}}$\\\\\\ Now Equality Hold When $\bf{4^{x.\sin^{-1}(x)}=4^{x.\cos^{-1}(x)}}$\\\\\\ So $\bf{x.\sin^{-1}(x)=x.\cos^{-1}(x)}$\\\\\\ So $\boxed{\boxed{\bf{x=0\;,\frac{1}{\sqrt{2}}}}}$