Finding root of an equation

Question

Find the number of real root in the equation

$2\cos \left ( \frac{x^{2}+x}{6} \right )=2^{x}+2^{-x}$

I got one root ,i.e. $\cos \left ( \frac{x^{2}+x}{6} \right )$

Is it correct??

Answer 1

Hint:

$2^x + 2^{-x} \geq 2$ by AM-GM, and $2\cos(\frac{x^2+x}{6})\leq 2$

Answer 2

By AM-GM $$2\cos \left ( \frac{x^{2}+x}{6} \right )=2^{x}+2^{-x}\geq2\sqrt{2^x\cdot2^{-x}}=2.$$ The equality occurs for $2^x=2^{-x},$ which gives $x=0$.

But, $$2\cos \left ( \frac{x^{2}+x}{6} \right )\geq2$$ gives $$2\cos \left ( \frac{x^{2}+x}{6} \right )=2,$$ which says that it's ewnough to check that $0$ is indeed the root.

Can you end it now?