Show that $x^2 -6 = \csc(x)$ has at least $3$ roots.

Question

I need to show that $$x^2 -6 =\csc(x)$$ has at least $3$ roots.

How can I show that? I tried by using mean value theorem...

Answer 1

Hint. Since RHS is a periodic function of $x,$ and is unbounded whenever $x=kπ,$ where $k\in\mathrm Z,$ then it follows that the curve defined by RHS will intersect this curve infinitely many times.

Answer 2

Let $$ f(x)=(x^2-6)\sin(x)-1. $$ Then that the equation has at least three roots is equivalent to that $f(x)$ has at least three roots. Clearly $f(x)$ is continuous in $(-\infty,\infty)$. Define $$ x_1=-\frac{3\pi}{2},x_2=-\pi,x_3=-\frac{\pi}{2},x_4=0 $$ and then $x_1<x_2<x_3<x_4$. It is easy to see \begin{eqnarray} f(x_1)&=&\frac{9\pi^2-27}{4}>0,\\ f(x_2)&=&-1<0,\\ f(x_3)&=&5-\frac{\pi^2}{4}>0,\\ f(x_4)&=&-1<0 \end{eqnarray} and hence there are three roots in $(x_1,x_2), (x_2,x_3),(x_3,x_4)$, respectively, by the IMVT.