Answer: D

Solution:

Let $f(\theta)=1-3\sin\theta+5\cos 3\theta.$ At $\theta=0,\tfrac\pi3,\tfrac{2\pi}3,\ldots,2\pi,$ its signs alternate $+,-,+,-,+,-,+.$ Therefore there is at least one root in each of the six intervening intervals.

At any root, $5\cos3\theta=3\sin\theta-1.$ Consequently $$\begin{aligned} &25\sin^2 3\theta-\cos^2\theta \\ &\quad =23+6\sin\theta-8\sin^2\theta>0. \end{aligned}$$ Hence $15|\sin3\theta|>3|\cos\theta|.$ In the interval $(\tfrac{k\pi}{3},\tfrac{(k+1)\pi}{3}),$ the sign of $f'(\theta)=-3\cos\theta-15\sin3\theta$ is therefore $(-1)^{k+1}$ at every root.

Thus every root in a given interval crosses the axis in the same direction. Two such roots would require an intervening crossing in the opposite direction, so each interval has exactly one root. There are $6$ solutions in all.

Thus, the correct answer is D.