Finding zeros in a goniometric function

Question

I need to find the zeros (roots?) of a function e.g. $$x-\sin(2x)$$ in the interval $[0,\pi]$ so that would make $$\sin(2x) = x$$

I've already found $0$ but I can't find a way to determine the other one

Answer 1

As $\sin 2x\approx 2x $ for $x\approx 0$ you know that $$ \sin(2x)>x $$ on some interval right of $x=0$. More precisely, by alternating series properties $$\sin (2x)\ge 2x-\frac{(2x)^3}6=x\left(2-\frac{4x^2}3\right)$$ for $|x|<1$, so that you can take $x_1=\frac34$.

For $x>1$ you know that $$\sin(2x)\le 1<x.$$

Combined this means that there is a root of the equation in $$ x\in\left[\frac34,1\right]. $$ Now check the derivatives over this interval to conclude that there is exactly one root because of monotonicity.