2017 AMC 12A Problem 15

For $0\lt x\lt\dfrac{\pi}{2}$ all three terms are positive, so $f(x)\gt0.$ For $\dfrac{\pi}{2}\lt x\lt\pi,$ set $s=\sin x\gt0$ and $t=-\cos x\gt0.$ Since $0\lt t\le1,$ $$\begin{aligned} f(x)&=s-2t-\dfrac{3s}{t} \\ &=-2t-s\left(\dfrac{3}{t}-1\right)\lt0. \end{aligned}$$ Thus no root occurs before $x=\pi.$ At $x=\pi,$ $f(\pi)=0+2(-1)+0=-2\lt0.$ At $x=\dfrac{5\pi}{4},$ $\tan x=1$ so $f=-\dfrac{\sqrt2}{2}+3\gt0.$ By the intermediate value theorem the smallest positive root lies in $\left(\pi,\dfrac{5\pi}{4}\right).$ Since $\pi\gt3$ and $\dfrac{5\pi}{4}\lt4,$ this interval sits inside $(3,4).$ Thus, the correct answer is D.