Number of roots of the equation $x\sin x-1=0$ for $x\in [0,2\pi]$

Question

Number of roots of the equation $x \sin x-1=0$ for $x\in [0,2\pi]$

My attempt is using the bisection method where I initially took $a=0$, $b=\frac{\pi}{2}$

as $f(a)\cdot f(c)<0$

We can proceed along the lines

we get one root $\in[0,\frac{\pi}{2}]$

Similarly we get another root $\in[\frac{\pi}{2},\pi]$

as $f(a)\cdot f(c)<0$ again

but no roots $\in[\pi,2\pi]$ as $f(a)\cdot f(c) \not< 0$

Hence we can concude we have two roots $\in[0,2\pi]$

Please tell me if I am on the correct lines. Ideas,solutions are appreciated.

And if possible tell me tricks to solve such types of transcendental equations.

The curve of xsinx =0 is given above. See that the ampltude of the curve goes on increasing as $(\frac{\pi}{2},\frac{5\pi}{2}...)$

This proves that if $x \in [0,2\pi]$, the function $xsinx$ can attain a value $=1$ only at two instances i.e on the +ve half of the first wave. Hence we are done .

This is a lot handwavy still gives us a correct answer I guess.

Answer 1

Here is the plot of f(x) = $x \sin(x) - 1$ for $0\le x \le 2\pi $.

If $f(a)f(c)\lt0$ there must be at least one root between $a$ and $c$ but there could be more!

Also if $f(a)f(c)\gt0$, it does not mean that there is no root between $a$ and $c$. In fact, you could have infinite number of roots there. There is no simple way to tell the number roots in general case.

In this particular case it is obvious that $x\sin(x) \le 0$ for $\pi \le x \le 2\pi$ so there's no root there.

For $0 \le x \le \pi$, the function $x\sin(x)$ goes from zero to zero and passes through $\pi/2$ for $x=\pi/2$. It means that the graph of the functon $x\sin(x)$ must cross line $y=1$ at least twice so you have at least two roots in this region.

Answer 2

Clearly 0 is not a root . See that $ x\sin x -1=0 \Rightarrow \sin x = 1/x $ when $x \neq 0$ . So you have to see that how many times the graphs of $\sin x$ and $x$ intersect each other in the interval $(0,2\pi]$. Now you draw both the graphs of $\sin x$ & $x$ and observe that they intersect twice when $ x \in (0,2\pi] $ . So there are exactly 2 roots of $ x\sin x -1=0$ in $[0,2\pi]$ .

Answer 3

Rewrite as

$$ x = \frac{1}{\sin x} = \csc x $$

It is then clear that there can be no solutions in $(\pi, 2\pi)$, since $\csc x$ is negative there, while $x$ itself is positive.

On the interval $(0, \pi)$, note that

• $x < \csc x$ as $x$ approaches $0$ from above
• $x > \csc x$ at $x = \pi/2$
• $x < \csc x$ as $x$ approaches $\pi$ from below
• $\csc x$ is concave upward across the entire interval (are you at the point in your studies where you can show this?)
• $x$ itself is a linear function

Therefore, $x = \csc x$ must have exactly two solutions in that interval. Since $x = 0$, $x = \pi$, and $x = 2\pi$ are not solutions, there can be only two solutions in the interval $[0, 2\pi]$.