Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 5 / b / Solution

Substitution into the circle gives $x^2{\sec }^{2}x=25$, or $x=5|\cos x|$ since $x\ge 0$. Necessarily $x\le 5$, and each admissible $x$ fixes a unique $y=x\tan x$.

On $[0,\pi /2)$, $x-5\cos x$ is strictly increasing from a negative value to a positive value, giving one root. On $(\pi /2,3\pi /2)$, the equation is $x+5\cos x=0$. This function is strictly convex, positive at the two interval ends and negative at $\pi$, giving exactly two roots.

There are none on the remaining possible interval $3\pi /2<x\le 5$. Put $\delta =x-3\pi /2>0$; then $\cos x=\sin \delta \le \delta$, soHence the number of simultaneous real solutions is $3$, as marked in the preceding graph.