How many real solutions does this equation have? $x^2 + \cos x = 15$

Question

How many real solutions does this equation have? $x^{2} + \cos(x) = 15$

Another task from an old exam I'm not sure how to solve!

The first thing that came to my mind was the well known and easy p-q-formula.

But I didn't know how to apply that formula if we got a cos. We cannot just write:

$$x_{1,2} = -\frac{\cos}{2}+- \sqrt{\left (\frac{\cos}{2} \right )^{2}+15}$$

Is there even one solution? I don't think.. We could get +- 14 as solution but never 15.

Very complicated task..

Answer 1

Let $f(x)=x^2+\cos(x)-15$. You have that $f'(x)=2x-\sin(x)$.

$$f'(x)=0\implies 2x=\sin(x)\implies x=0.$$ Moreover $f''(0)=1>0$, and thus $f(0)$ is a minimum. Since $f(0)=-14$ and that $\lim_{x\to \pm \infty }f(x)=+\infty $, you will have only two solution at the equation $f(x)=0$.

Answer 2

Notice that once $|x| > 1/2$ the function $x\mapsto x^2 + \cos(x)$ is $1-1$.

Answer 3

The function is even and we can limit the study to positive solutions.

As $-1\le\cos x\le 1$, the roots will be "bracketed" by those of $x^2+t=15$ with $t\in[-1,1]$, hence

$$x=\sqrt{15-t}\in[\sqrt{14},4].$$

In this range, the derivative $2x-\sin x$ certainly remains positive and the function is strictly increasing, hence has a single root.