15. Given $f(x)=\sin x+2 \cos x+3 \tan x$, where $x$ is in radians. Then the interval ( ) contains the smallest positive real number $x$ such that $f(x)=0$.
(A) $(0,1)$
(B) $(1,2)$
(C) $(2,3)$
(D) $(3,4)$
(E) $(4,5)$
15. D.

When $x \in(0,1)$, $f(x)$ is monotonically increasing, and $f(0)=2$, so there is no zero point in the interval $(0,1)$;
When $x \in\left(1, \frac{\pi}{2}\right)$, $f(x)$ is monotonically increasing, and from the previous part, $f(1)>0$, so there is no zero point in the interval $\left(1, \frac{\pi}{2}\right)$;
When $x \in\left(\frac{\pi}{2}, 3\right)$, $f(x)$ is monotonically increasing, and $f(3)0,
\end{array}
$$

there is a zero point in the interval $\left(\pi, \frac{5 \pi}{4}\right)$.
Since $\frac{5 \pi}{4}<4$, therefore, $f(x)$ has a zero point in the interval $\left(\pi, \frac{5 \pi}{4}\right)$.