How many roots does $(x+1)\cos x = x\sin x$ have in $(-2\pi,2\pi)$?

Question

So the nonlinear equation that I need to find the number of its roots is $$(x+1)\cos x = x\sin x \qquad \text{with } x\in (-2\pi,2\pi)$$

Using the intermediate value theorem I know that the equation has at least one root on this interval, and if I use drawing I see that $x\sin x$ and $(x+1)\cos x$ intersect in three points, but from drawing I can't know if they might intersect again somewhere.

And the problem is that the number of zeroes is definitely not 3, the options are 4, 5, 6, 7 based on my textbook. I tried the Fixed point method but $\{x\}$ didn't converge, either my starting point or the function I chose were inappropriate. Can you help?

Answer 1

From the hint given by @zkutch, it is evident from the graph that the equation has five roots when $x\in[-2\pi,2\pi]$. As suggested by @Claude Leibovici I've posted the original graph which is indeed more nice than the second one. However, students are more familiar with the second one. Third graph if necessary. :-)

Answer 2

First we can manipulate the expression given:

$(x+1)\cos x= x\sin x$

divide both sides by $x\cos x$

Yielding: $(\frac{x+1}{x})= \tan x $

The Inverse tangent of both sides yields:

$\operatorname{arctan}(\frac{x+1}{x})=x$

We can state without requiring a proof, although can be proved, that: $\frac{d}{dz} \operatorname{arctan}(z) = \frac{1}{1+z^2}$

In our example we could let $z=\frac{x+1}{x}$

So: $\frac{d}{dx}\operatorname{arctan}(z)= (\frac{d}{dz}\operatorname{arctan}(z)\cdot \frac{dz}{du})$ By the Chain Rule !

Right Hand Side (RHS) of the above expression yields $\frac{1}{1+z^2} \cdot \frac{-1}{x^2}$

Substitute value of $z$ in terms of $x$ into the RHS to yield an overall equation:

$\frac{d}{dx}\operatorname{arctan}(z)= \frac{1}{1+(\frac{x+1}{x})^2}\cdot\frac{-1}{x^2}$ $= \frac{-1}{x^2+(x+1)^2}$

$ \therefore , \operatorname{arctan}(z)=\int \frac{-1}{x^2+(x+1)^2}$

and as: $\operatorname{arctan}(z)= x$. Then:

$x=\int \frac{-1}{x^2+(x+1)^2}$

Hopefully you can solve this integral now to find all the solutions for x