URGENT A is a point on the circumference of a cirlce.Chords AB and AC divide the area of the circle into three equal parts.If the angle BAC is the root of the equation,f(x)=0 then find f(x)

let the radius of the circle be r and the centre of the circle be O.
given: the chord AB and AC divide the circle into three equal parts.
let the angle BAC be x.
$$\begin{gathered} ∆OAB\cong ∆OAC \\ [OA=OB=OC [radius] \\ AC = AB \end{gathered}$$
three corresponding sides are equal therefore triangles are congruent.
$\angle OAB=\angle OBA=\angle OCA=\angle OAC=x/2$
$\angle AOB=\mathrm{\pi }-(\angle OAB+\angle OBA)=\mathrm{\pi }-\mathrm{x}$
the area of the shaded region= area of the sector OAB - area of triangle OAB
$$\begin{gathered} {\mathrm{\pi r}}^2*\frac{\mathrm{\pi }-\mathrm{x}}{2\mathrm{\pi }}-\mathrm{area}(∆\mathrm{OAB}) \\ ={\mathrm{\pi r}}^2*\frac{\mathrm{\pi }-\mathrm{x}}{2\mathrm{\pi }}-\frac{1}{2}*\mathrm{rsin}\frac{\mathrm{x}}{2}*\left(2\mathrm{rcos}\frac{\mathrm{x}}{2}\right) \\ ={\mathrm{\pi r}}^2*\frac{\mathrm{\pi }-\mathrm{x}}{2\mathrm{\pi }}-\frac{{\mathrm{r}}^2}{2}\mathrm{sinx} \end{gathered}$$
since the area of the shaded region is the one-third of the area of the circle.
therefore
$$\begin{gathered} {\mathrm{\pi r}}^2*\frac{\mathrm{\pi }-\mathrm{x}}{2\mathrm{\pi }}-\frac{r^2}{2}\sin x=\frac{1}{3}*{\mathrm{\pi r}}^2 \\ \pi *\frac{\mathrm{\pi }-\mathrm{x}}{2\mathrm{\pi }}-\frac{1}{2}\sin x=\frac{1}{3}*\pi \\ \frac{\mathrm{\pi }}{2}-\frac{x}{2}-\frac{\sin x}{2}=\frac{\mathrm{\pi }}{3} \\ \frac{x}{2}+\frac{\sin x}{2}=\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{3} \\ \frac{x}{2}+\frac{\sin x}{2}=\frac{\mathrm{\pi }}{6} \\ x+\sin x=\frac{\mathrm{\pi }}{3} \\ x+\sin x-\frac{\mathrm{\pi }}{3}=0 \end{gathered}$$
therefore angle BAC is the root of the function $f\left(x\right)=x+\sin x-\frac{\mathrm{\pi }}{3}$