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Exercise 369
Let $f(x)=\sin x+3\cos x+2\tan x$. In what unit interval does the smallest positive $x$ with $f(x)=0$ lie?
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Solution
  1. On $\left(0,\frac\pi2\right)$ all terms are positive. On $\left(\frac\pi2,\pi\right]$, $\sin x+2\tan x=\sin x\left(1+\frac2{\cos x}\right)\le0$ (as $\frac2{\cos x}\le-2$) and $3\cos x<0$, so $f<0$. No roots there.
  2. On $\left(\pi,\frac{3\pi}2\right)$, $f'(x)=\cos x-3\sin x+\frac2{\cos^2x}\ge-1+0+2>0$, so $f$ is increasing and has at most one root. Now $f(4)\approx-0.757-1.961+2.316<0$ and $f(4.5)\approx-0.978-0.633+9.27>0$, so the root lies in $(4,4.5)$.
  3. The smallest positive root lies in $\boxed{(4,5)}$.
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