AMC12 2017 A

Answer (D): For $0<x<\frac{\pi}{2}$ all three terms are positive, and $f(x)$ is undefined when $x=\frac{\pi}{2}$. For $\frac{\pi}{2}<x<\frac{3\pi}{4}$, the term $3\tan x$ is less than $-3$ and dominates the other two terms, so $f(x)<0$ there. For $\frac{3\pi}{4}\le x<\pi$, $|\cos(x)|\ge|\sin(x)|$ and $\cos x$ and $\tan x$ are negative, so $\sin x+2\cos x+3\tan x<0$. Therefore there is no positive solution of $f(x)=0$ for $x<\pi$. Because the range of $f$ includes all values between $f(\pi)=-2<0$ and $f\!\left(\frac{5\pi}{4}\right)=-\frac{3}{2}\sqrt{2}+3>-1.5\cdot1.5+3>0$ on the interval $\left[\pi,\frac{5\pi}{4}\right]$, the smallest positive solution of $f(x)=0$ lies between $\pi$ and $\frac{5\pi}{4}$. Because $\pi>3$ and $\frac{5\pi}{4}<4$, the requested interval is $(3,4)$.