Number of real solutions $2\cos(\frac{x^2+x}{2})=2^x+2^{-x} $

Question

The number of real solutions of $2\cos(\frac{x^2+x}{2})=2^x+2^{-x} $ is

(1) 0

(2) 1

(3) 2

(4) infinitely many .

My work :

$$ 1\geq \cos\left(\frac{x^2+x}{2}\right)=\frac{2^x+2^{-x} }{2}\geq 1 \qquad \text{by (AM-GM).} $$

So $\frac{x^2+x}{2}=2n\pi$ for all $n\in \mathbb{Z}$ . Now discriminant $=1+2n\pi$ is always positive for $n\geq0$ . But the equation is a quadratic so it has only two solution . Hence the answer must be 2 .

PS: I'm aware that the this problem is already on the site but i posted as there were no complete solution .

See also : https://math.stackexchange.com/questions/2266973/cosy-x-frac1x-possible-for-any-values-of-y/2267090#2267090 — lab bhattacharjee, May 09 '17 at 15:38

score 2 · Accepted Answer · answered May 09 '17 at 15:15

2

Your inequality implies to get a solution you must have $$ 2^x + 2^{-x} = 2 $$ which only happens when $2^x=1$, so there is only one solution.

answered May 09 '17 at 15:15

gt6989b

54,422

score 1 · Answer 2 · answered May 09 '17 at 15:14

1

Answer is 1. You wrote that it should be definitely $2^x + 2^{-x} = 2$. Let $y = 2^x$, so we are intereseted in $y^2 - 2y + 1 = (y - 1)^2 = 0$. The only solution comes when $y = 1$ or $x = 0$.

answered May 09 '17 at 15:14

Symmetrical

467

score 1 · Answer 3 · answered May 09 '17 at 15:15

1

hint

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

$$2^x+2^{-x}\geq 2$$

the root must satisfy

$$2^x+2^{-x}=2$$ which gives $x=0$. the unique root.

answered May 09 '17 at 15:15

hamam_Abdallah

62,951

Number of real solutions $2\cos(\frac{x^2+x}{2})=2^x+2^{-x} $

3 Answers3