Number of real roots of the equation $2\cos(x) =(2^x+2^{-x})/2$

Question

What's the number of real roots of the equation $2\cos(x) =(2^x+2^{-x})/2$?

If any questions ask about real root what is the main thing to check first, and most important?

Any particular way to handle these type of questions?

Answer 1

i doubt if this has a closed form solution; numerical solution is $\pm 0.9206433.$

Answer 2

You get the existence of two solution by mean value theorem, we have $$2\cos(0)-\frac{2^0 +2^{-0}}{2}=2-1=1>0$$ but for $|x|$ large enough it is clear that $$2\cos(x)-\frac{2^x +2^{-x}}{2}<0$$ because $2^{|x|}$ goes to infinity. The functions are continous, so we get a positive and a negative real solution.

You can also show that there are only two solutions because $\frac{2^x +2^{-x}}{2}$ is monotonous on $(0,\infty)$ and for $x=2$ it's already greater than $2=\max_x 2\cos(x)$.

Answer 3

It suffices to consider the interval $(0,\infty)$, as if $x$ is a solution, also $-x$ is.

The function $x\mapsto (2^x-2^{-x})/2$ is increasing and convex, by examining its derivatives.

There is no solution with $x>\pi/2$, because

• for $\pi/2<x<\pi$ we have $2\cos x<0$ and $(2^x-2^{-x})/2>0$

• for $x\ge\pi$, $2\cos x\le2$, whereas $(2^x-2^{-x})/2>(2^3-2^{-3})/2>4$.

So we can restrict the search to the interval $[0,\pi/2]$, where the function $x\mapsto 2\cos x$ is concave and decreasing.

Thus if a solution exists it is unique.

Since $2\cos0=2$ and $(2^0+2^{-0})/2=1$, whereas $(2^{\pi/2}+2^{-\pi/2})/2>0=2\cos(\pi/2)$, a solution exists.