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

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 $[0, 2\pi]$.
