Solution of trigonometric equation $0 = x\cos(x)+2$

Question

I'm having a hard time finding the solution of the following equation:

$$0 = x\cos(x)+2$$

This is part of showing that $f(x):= x^2e^{\sin(x)}$ has 3 extremas in $[-\frac{\pi}{2},\frac{\pi}{2}]$.

Answer 1

If:

$$y(x)=x^2 e^{\sin x}$$

then:

$$y'=xe^{\sin x}(x\cos x + 2)$$

Now if you put $y'=0$ we have:

$$xe^{\sin x}(x\cos x + 2)=0$$

Note that $e^{\sin x}$ > 0 so we get:

$$x(x\cos x +2)=0$$

...which means that:

$$x = 0 \ \ \lor \ \ (x \cos x +2) =0$$

For all values in the interval $x\in [0,\pi/2]$ obviously $x\cos x+2>0.$

For $x\in[-\pi/2, 0)$, $|x \cos x| \le|x|<\pi/2 < 2$ so in this interval we aslo have $x\cos x+2>0.$

This leaves us with only one possibility for an extremal value in $[-\pi/2,\pi/2]$ and that is $x=0$.

The plot of $y(x)$ just confirms that you cannot have 3 extremas in the specified interval: