# Find number of solutions of the equation $2\sin|x|-x=0$

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$$?$$

• Draw a graph of $y=2\sin|x|$ and then draw a graph of $y=x$ on the same axes. Jan 9, 2023 at 12:30
• There are only two solutions that can be seen graphically. Now you have to prove it properly. Jan 9, 2023 at 12:40
• @Lelouch that's what I want to know....the method Jan 9, 2023 at 12:52
• You can use the intermediate value theorem on each interval where the function is monotonic. Jan 9, 2023 at 12:52
• $x=0$ is a trivial solution. Now consider cases $x <0, x>0$ and apply regular analysis using derivatives to determine local extrema and intervals of increasing/decreasing. Jan 9, 2023 at 12:52

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.

If $$\,x\leqslant-\pi\,,\;$$ it results that $$\,2\sin|x|-x\geqslant-2-x\geqslant-2+\pi>0\,.$$

If $$\,x\!\in\,]\!-\!\pi,0[\,,\;$$ it results that $$\,2\sin|x|-x>0-x>0\,.$$

If $$\,x>\pi\,,\;$$ it results that $$\,2\sin|x|-x\leqslant2-x<2-\pi<0\,.$$

Consequently, there is not any solution of the equation $$\,2\sin|x|-x=0\;\;$$ in $$\;\;]-\infty,0[\,\cup\,]\pi,+\infty[\,.$$

Let $$\;f:[0,\pi]\to\Bbb R\;$$ be the function defined as

$$f(x)=2\sin|x|-x=2\sin x-x\quad$$ for all $$\,x\!\in[0,\pi]\,.$$

The function $$\,f\,$$ is continuous on $$\,[0,\pi]\,$$ and differentiable on $$\,]0,\pi[\,,\,$$ moreover, $$\;f’(x)=2\cos x-1\quad$$ for all $$\,x\!\in\,]0,\pi[\,.$$

Since $$\;f’(x)>0\;\;$$ for all $$\,x\!\in\,]0,\pi/3[\;\;$$ and $$\;f’(x)<0\;\;$$ for all $$\,x\!\in\,]\pi/3,\pi[\;\;,\;\;$$ it follows that the function $$\,f\,$$ is increasing on $$\,[0,\pi/3]\,$$ and is decreasing on $$\,[\pi/3,\pi]\,.$$

Furthermore ,

$$f(0)=0\;,\;\;f(\pi/3)=\sqrt3-\pi/3>0\;,\;\;f(\pi)=-\pi<0\,.$$

Consequently, there exists only one real number $$\,c\!\in\,]\pi/3,\pi[\;$$ such that

$$f(x)>0\quad$$ for all $$\;x\!\in\,]0,c[\;\;,$$

$$f(c)=0\;\;,$$

$$f(x)<0\quad$$ for all $$\;x\!\in\,]c,\pi]\;.$$

Hence, the equation $$\;2\sin|x|-x=0\;$$ just has two solutions that are $$\;x=0\;$$ and $$\;x=c\;$$ where $$\;c\!\in\,]\pi/3,\pi[\,.$$

There are only two solutions. The question didn't ask about the numerical values for a specific reason, there is no straightforward way of obtaining them (apart from the trivial solution of x=0). It is geared towards the 'reading graphs' side of thinking. Your friends made an error in that they probably looked at the graph of $$\sin x$$, not the $$\sin \lvert x \rvert)$$. You can go to desmos.com for example and convince yourself, graph $$\sin \lvert x \rvert$$ and separately $$x/2$$, the solutions are intersections of those graphs (i.e. the places where they have the same value for the same argument).

This is the graph of Equation. This is not the solution. But I think it might give you an idea about the question.