Number of roots between $–\pi$ and $\pi$ of the equation $\frac{2}{3} \sin x\cdot x=1$ is

Question

Number of roots between $–\pi$ and $\pi$ of the equation $\frac{2}{3} x\sin x = 1$ is

My Work:-

Plotting $y = \sin x$ & $xy = \frac{3}{2}$

I am getting 4 roots. How to conclude it without plotting graph?

for reference https://www.desmos.com/calculator/jap0g2ju31

Answer 1

The curve $y=\sin x$ is concave down on $[0,\pi]$ while the curve $y=3/(2x)$ is concave up. Thus they can intersect at most twice on $[0,\pi]$. Since $\sin0=0\lt\infty=\lim_{x\to0}3/(2x)$, $\sin(\pi/2)=1\gt3/\pi=3/(2\pi/2)$, and $\sin(\pi)=0\lt3/(2\pi)$, the curves do intersect twice on $[0,\pi]$. By symmetry, they intersect twice on $[-\pi,0]$ as well, hence there are four roots in all.

Answer 2

Hint

As the map $f(x) =\frac{2}{3}\sin x \cdot x$ is even, it is sufficient to prove that there are exactly two solutions on $(0,\pi]$ as $0$ is not a solution.

Then $a$ is solution if and only if $\frac{2}{3} \sin a =\frac{1}{a}$.

$g(x) = \frac{2}{3} \sin x - \frac{1}{x}$ is increasing (difference of an increasing and a decreasing map) on $(0,\pi/2]$ from $-\infty$ to $2/3-2/\pi >0$ hence have a unique solution on this interval.

Remains to find a simple argument on why there is also only one solution on the interval $[\pi/2, \pi]$. Proving that one exists being easy as $\sin \pi=0$.