There is an other equation here: $$\sin{x}-x\cos{x}=0$$ Range for x : $[0,\frac{3\pi}{2}]$
Now we want to write the equation $f(x)$ like $h(x)=g(x)$ : $$\sin{x}=x\cos{x}$$ It is in a way that we know how to draw h and g functions diagram. Then we draw the $h$ and $g$ function diagrams and find the common points of them. So it will be number of the $f(x)$ roots that here is the equation mentioned top. Actually now my problem is with drawing the first equation's diagram I want you to draw its diagrams step by step. Please help me with it! Just help me drawing them.
From $f(x):=\sin x- x\cos x$ we compute $f'(x)=x\sin x$. This shows that $f$ is strictly increasing for $0\leq x\leq \pi$ and strictly decreasing for $\pi\leq x\leq{3\pi\over2}$. As $f(0)=0$, $f(\pi)=1$, and $f\bigl({3\pi\over2}\bigr)=-1$ we can conclude that there are exactly $2$ roots in the interval $\bigl[0,{3\pi\over2}\bigr]$, namely $0$ and a single root in $\>\bigl]\pi,{3\pi\over2}\bigr[\>$.
A "diagram" can only visualize what is written here, but cannot be considered as a proof per se. (There is no ISO-certificate for diagrams.)