Find number of solutions of the equation $$2\sin|x|-x=0$$
This was asked in asked in my mock exam. My two friends said that the answer was $3$ justifying it from the theory of graphs. I just gave $1$ as the answer as I am not well versed with graphs. But wolframaplha is showing that there are only two solutions of the above equation one being $0$ and the other is $\approx 1.89$
Who is correct and who isn't $?$ How to find their number$?$ Should I take help from graphs or is there some method in calculus for this$?$
Best Answer
The proper way to study the number of solution is to use the intermediate value theorem on each interval where the function is monotonic.
First of all, you can restrict the study to the interval $[-2,2]$ since $-1< \sin(|x|) < 1$ and thus there is no solution outside $[-2,2]$. It is not always needed to restrict the interval of study but here the function contains a periodic term and therefore it has multiples irrelevant local extrema. It is very convenient to restrict to a small interval to avoid studying theses local extremas.
The derivative of the function $f(x) = 2 \sin(|x|)-x$ is $f'(x) = \frac{2 x \cos(|x|)}{|x|} - 1$.
The only solution to $f'(x)=0$ is when $x=\frac{\pi}{3}$ (in the interval $[-2,2]$). Also note that $f$ is not differentiable at $x=0$ therefore we have to be careful about this point.
We have $$f(-2) \approx 3.82 $$ $$f(0)= 0$$ $$f(\frac{\pi}{3}) \approx 0.68 $$ $$f(2) \approx -0.18$$
On each of the intervals $[-2, 0], [0, \frac\pi3]$ and $[\frac\pi3, 2]$, the function is continuous and strictly monotonic. Therefore we can use the intermediate value theorem to show there is possibly a root, and the strict monotonicity will give the uniqueness of the root inside the interval if the root exists.
There is exactly one root inside the interval $[-2, 0]$ because $f$ is continuous, strictly monotonic and $f(-2)f(0) \leq 0$ (opposite signs at the boundaries). In this particular case we know the root is $x=0$ because $f(0)=0$.
Similarly, there is exactly one root inside the interval $[0, \frac\pi3]$ and we know it's still $0$ because the root is unique.
There is exactly one root in the interval $[\frac\pi3, 2]$ because $f$ is continuous, strictly monotonic, $f(\frac\pi3) > 0$ and $f(2) < 0$.
In the end, we have two roots : $x=0$ is a root, and there is another root in the interval $[\frac\pi3, 2]$. You can approximate the unknown root using numerical approximations but there is no exact value for it.
I hope it's clear enough to give you an idea of how to solve such problems.