Number of real roots of an equation

Question

What is the number of real roots of the equation

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

How to solve this kind of problems. Any general methods ??

http://math.stackexchange.com/questions/380896/how-to-find-the-number-of-real-roots-of-the-given-equation — lab bhattacharjee, May 04 '13 at 12:06

score 2 · Answer 1 · answered May 04 '13 at 12:15

2

Hint: Minimum value of RHS =2.

answered May 04 '13 at 12:15

score 1 · Answer 2 · answered May 04 '13 at 12:17

By $AM-GM$, $$\frac{2^x+2^{-x}}{2} \geq \sqrt{2^x2^{-x}} = 1$$ $$\implies 2^x+2^{-x} \geq 2$$

with equality only when $2^x = 2^{-x}$, i.e. $x=0$.

Also, $$2\cos(\frac{x^2+x}{6}) \leq 2$$

So the only solution is when $LHS=RHS=2$. The only olution for $RHS=2$ is $x=0$, which also turns out to be a solution for $LHS$.

Hence there is exactly one real root, $x=0$.

2 Answers2