A geometry problem by karan tatariya

$2x=3π(1-\cos x)$ $\Rightarrow \sin^2 \frac{x}{2}=\frac{x}{3π}$ Putting $\frac{x}{2}=X$ $\Rightarrow \sin^2 X=\frac{2X}{3π}$ Now visualization of graph of these two functions is hepful. One might think that there are three solutions one at $X=0$ the other at $X= \frac{3π}{4}$ & the third one at $X=\frac{3π}{2}$ .

But actually it's not.

On examining the derivative of $\sin^2 X$ , the slope is zero at $X=0$ & increases further thereby meeting the line $Y= \frac{2X}{3π}$ two times near the origin. The same scenario happens at $X=\frac{3π}{2}$ . Hence there are a total of $2+1+2=\boxed5$ solutions of the equation.

The graph of 3 $\pi$ (1- $\cos x$ ) and 2x can be easily visualised as below:

graphs

From here there is clearly one intersection at $\frac {3\pi} {2}$ .

Near the points 0 and 3 $\pi$ there will be 2 intersection points each.

Because slope of 3 $\pi$ (1- $\cos x$ ) at both 0 and 3 $\pi$ is 0.

Slope of 2x is 2 at both of these points, hence the line 2x must cross over at 2 other points (after 0 and before 3 $\pi$ ).