Prove that f(x)=0 has exactly 3 solutions I need to prove that tan(x)=10x has exactly 3 solutions in $[-\pi/2,\pi/2]$:
I declared $f(x)=\tan(x)-10x$

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*I have proved that $f(x)=0$ has at least 3 solution using the fact that it's odd and continuous function.


*Now I need to prove that there can't be 4 solutions, I remember a rule that says if $f(x)=0$ in four points then f'(x)=0 in three point in $[-\pi/2,\pi/2]$ but how may I continue from here?
Note: $f'(x)=\frac1{\cos^2 x}-10$

On $[0, \frac {\pi} 2]$ there is a unique point where $cos^{2}(x)=\frac  1{10}$. [$\cos x$ decreases strictly fron $1$ to $0$ in this interval].   Same is true in $[- \frac {\pi} 2, 0]$. So there are exactly two points in $[- \frac {\pi} 2,\frac {\pi} 2]$ where $f'(x)=0$.
