**Problem:** Show that there is exactly one value of \(x\) that satisfies the equation:

\(2 cos^2(x^3+x)=2^x+2^{-x}\)

**Solution:** We know that \(cos \;x \leq 1\) for all \(x \in I\!R\)

\(=> cos(x^3 + x)\leq 1\)

\(=> cos^2(x^3 + x)\leq 1\)

\(=> 2cos^2(x^3 + x)\leq 2\)

Now consider \(2^x\) and \(2^{-x}\). By AM-GM inequality we have,

\(2^x+2^{-x}\geq 2\)

So \(2 cos^2(x^3+x)=2^x+2^{-x}\), only when \(2 cos^2(x^3+x)=2=2^x+2^{-x}\).

That means \(2^x+2^{-x} = 2, => x= 0\). So \(x=0\) being the only solution.

and \(x=0\) also satisfies \(2 cos^2(x^3+x)=2\)

Thus there is exactly one solution.

Hence Proved.