Trigonometric equation: $2\arcsin \left(\frac{2x}{1+x^2}\right)- \pi x^3 = 0$

Question

The number of solutions of the equation $$2\arcsin \left(\dfrac{2x}{1+x^2}\right)- \pi x^3 = 0$$ is?

Let $x= \tan \theta$

$\implies \sin 2\theta = \sin(\dfrac \pi 2 \tan^3\theta)$

I had to delete the rest of my attempt because it was totally wrong. What are the methods to solve this problem?

Answer 1

As J.G. noted in comments, there are no solutions for $|x|\gt1$. This is because the range of the arcsine function is $[-\pi/2,\pi/2]$, so the left hand side, $2\arcsin\left(2x\over1+x^2\right)$, is never greater than $\pi$ in absolute value, while $\pi|x^3|\gt\pi$ if $|x|\gt1$. We'll come back to this in a moment.

Now after substituting $x=\tan\theta$, the left hand side becomes

$$2\arcsin\left(2\tan\theta\over1+\tan^2\theta\right)=2\arcsin\left(2\sin\theta\cos\theta \right)=2\arcsin(\sin2\theta)$$

It's tempting to "cancel" the the inverse trig function, leaving just $4\theta$. But that's only correct for $|\theta|\le{\pi\over4}$. However, that is exactly the domain we need in order to have $|x|\le1$, since $-1\le\tan\theta\le1$ for $-\pi/4\le\theta\le\pi/4$. So we can go ahead and write the equation as

$$4\theta=\pi\tan^3\theta$$

with the understanding that we only care about solutions with $-\pi/4\le\theta\le\pi/4$.

Three solutions jump off the page: $\theta=0$, $\theta=\pi/4$, and $\theta=-\pi/4$. The question is. are there any others? For this it helps to look at the function $f(\theta)=4\theta-\pi\tan^3\theta$, for which

$$f'(\theta)=4-3\pi\tan^2\theta\sec^2\theta$$

and $$f''(\theta)=-3\pi(2\tan\theta\sec^4\theta+2\tan^3\theta\sec^2\theta)=-6\pi\tan\theta\sec^2\theta(2\sec^2\theta-1)$$

It's clear that $f''(\theta)\lt0$ for $0\lt\theta\lt\pi/4$, and easy to see that $f'(0)=4\gt0$ while $f'(\pi/4)=4-6\pi\lt0$. Consequently there are no other solutions in $(0,\pi/4)$, hence, by symmetry, in $(-\pi/4,0)$ either.

To sum things up, $x=0$, $x=1$ and $x=-1$ are the only three solutions to

$$2\arcsin\left(2x\over1+x^2\right)-\pi x^3=0$$

Answer 2

Shouldn't the third line be $ 2\theta = \dfrac{\pi}{2}\tan^3\theta $?

The arcsin cancels out the sin.

Wolfy says that the roots in $[-\pi/2, \pi/2]$ are $0$ and $± 0.785398163397448...$.

Answer 3

Consider $$ f(x)=\arcsin\frac{2x}{1+x^2} $$ Then $$ f'(x)=\frac{1}{\sqrt{1-\dfrac{4x^2}{(1+x^2)^2}}}\frac{2(1+x^2)-4x^2}{(1+x^2)^2} =\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2} $$ Therefore $$ f'(x)=\begin{cases} \dfrac{2}{1+x^2} & |x|<1 \\[6px] -\dfrac{2}{1+x^2} & |x|>1 \end{cases} $$ which implies $$ f(x)=\begin{cases} c_--2\arctan x & x<-1 \\[6px] c_0+2\arctan x & -1\le x\le 1 \\[6px] c_+-2\arctan x & x>1 \end{cases} $$ and it's easy to see that $c_0=0$, $c_-=-\pi$ and $c_+=\pi$.

Thus the equation can be split into three cases: $$ \begin{cases} -2\pi-4\arctan x-\pi x^3=0 \\[4px] x<-1 \end{cases} \qquad \begin{cases} 4\arctan x-\pi x^3=0 \\[4px] -1\le x\le 1 \end{cases} \qquad \begin{cases} 2\pi-4\arctan x-\pi x^3=0 \\[4px] x>1 \end{cases} $$

The graph suggests that the solutions are only $-1$, $0$ and $1$ (which are indeed solutions).

Let's consider $\alpha(x)=-2\pi-4\arctan x-\pi x^3$ for $x\le-1$; then $$ \alpha'(x)=-\frac{4}{1+x^2}-3\pi x^2=-\frac{4+3\pi x^2+3\pi x^4}{1+x^2} $$ Note that the discriminant of $3\pi t^2+3\pi t+4$ is $9\pi^2-48\pi=3\pi(3\pi-16)<0$ so $\alpha$ is decreasing. As $\alpha(-1)=0$, there are no solutions of your equation for $x<-1$. Similarly, there are no solutions for $x>1$.

In the case $-1\le x\le 1$, the function $\beta(x)=4\arctan x-\pi x^3$ is odd, so we can study it over $[0,1]$. We have $$ \beta'(x)=\frac{4}{1+x^2}-3\pi x^2=-\frac{3\pi x^4+3\pi x^2-4}{1+x^2} $$ The derivative vanishes only for $$ x=\sqrt{\frac{-3\pi+\sqrt{9\pi^2+48\pi}}{6\pi}}\approx 0.5668 $$ Thus $0$ and $1$ are the only solutions on the interval $[0,1]$; by symmetry, $0$ and $-1$ are the only solutions on the interval $[-1,0]$.

It's not needed to look for an approximation: the following set of inequalities are equivalent to each other: \begin{gather} \sqrt{\frac{-3\pi+\sqrt{9\pi^2+48\pi}}{6\pi}}<1 \\[6px] \frac{-3\pi+\sqrt{9\pi^2+48\pi}}{6\pi}<1 \\[6px] -3\pi+\sqrt{9\pi^2+48\pi}<6\pi \\[6px] \sqrt{9\pi^2+48\pi}<9\pi \\[6px] 9\pi^2+48\pi<81\pi^2 \\[6px] 48<72\pi \\[6px] 2<3\pi \end{gather} and the last one is true.

Answer 4

Let $$ f(x)=2\arcsin\frac{2x}{1+x^2}-\pi x^3. $$ Then $f'(x)=-3\pi x^2-\frac{4}{1+x^2}<0$ if $|x|>1$, namely $f(x)$ is decreasing in $(-\infty,-1)$ and $(1,\infty)$. Also $$ f'(x)=-3\pi x^2+\frac{4}{1+x^2}, |x|<1$$ and $$ f''(x)=-\frac{2x[4+3\pi(x^2+1)^2]}{(x^2+1)^2}, |x|<1. $$ So $f''(x)>0$ if $x\in(-1,0)$ and $f''(x)<0$ if $(0,1)$ and hence $f(x)$ is strictly concave in $(-1,0)$ and strictly convex in $(0,1)$. Note that $f(-1)=f(0)=f(1)=0$. Thus $f(x)=0$ only has three roots $x=-1,0,1$.