Answer
a. see explanations.
b. only one zero.
Work Step by Step
a. Given $g(t)=sin^2t-3t$, we have $g'(t)=2sin(t)cos(t)-3=sin(2t)-3$. Because $-1\leq sin(2t)\leq 1$, we have $g'(t)\lt0$ over its entire domain; thus the function decreases on every interval in its domain.
b. Letting $f(t)=sin^2t-3t-5$, we have $f'(t)=sin(2t)-3\lt0$; thus $f(t)$ decreases over $(-\infty,\infty)$. Test $f(0)=-5\lt0$ and $f(-\frac{5\pi}{3})=sin^2(-\frac{5\pi}{3})+5\pi-5\gt0$, there will be only one zero in the interval of $(-\frac{5\pi}{3},0)$ as shown in the figure.
