Show that $x^3+2x=\cos x$ has exactly one solution in the interval $[0,\pi/2]$

Question

Show that $x^3+2x=\cos x$ have exactly one solution in the interval $[0,\pi/2]$.

Don't even know where to start.

Thanks in advance

Answer 1

Consider $f(x)=x^3+2x-\cos x$. Note that roots of $f$ are solutions to your original equation. Apply the Intermediate Value Theorem to show that it has a root in $(0,\pi)$.

Proceed by contradiction. Suppose $f$ has at least 2 roots $a,b$ in $(0,\pi )$. Apply Rolle's Theorem to deduce the existence of a $c\in (0,\pi)$ such that $f'(c)=0$. However, $f'(x)=3x^2+2+\sin x\ge 3x^2 + 1>0$, and there's the contradiction.

Answer 2

Show that $f(x)=x^3+2x-\cos x$ is strictly increasing in $[0,\pi/2]$.

$f'(x)=3x^2+2+\sin x>0$ for all $x\in[0,\pi/2]$, because $x^2\ge 0$ and $\sin x\ge 0$.

Now show that $f(0)<0$ and $f(\pi/2)>0$. Your result follows from the Intermediate Value Theorem.