This problem is to be solved within $3$ minutes, without a calculator.
My approach is bad:
The least positive solution of $\sin(x)=0$ is $\pi$,
The least positive solution of $\sin(x)+\frac{1}{2}\sin(2x)=0$ is "slightly" less then $\pi$,
When $k$ is "large", the terms approaches $0$. So I choose option (E). In fact, it is the correct answer.
I think a numerical method is to be used. I do not know how.
The word "slightly" and the word "large" are not technical here.
Please help me to solve this problem. Thanks in advance.
Suppose that you look at the non trivial positive solutions ($0$ and $\pi$) $x_n$ of the equation $$S_n(x)=\sum_{k=0}^{n}\frac{\sin(2^k x)}{2^k}=0$$
For $n=2$ (for $n<2$, only trivial roots) $$S_2=\sin (x)+\frac{1}{2} \sin (2 x)+\frac{1}{4} \sin (4 x)=\sin (x) \left(2 \cos ^3(x)+1\right)\implies x_2=\cos ^{-1}\left(-\frac{1}{\sqrt[3]{2}}\right)$$ This is the lower bound of all $x_n$. In other words $$\frac {4}5 \pi < x_n < \pi $$ So, option $(E)$ is the correct one.