Calculate minimum and maximum and zeros fo function 3sinx.
Thanx, by.
As the amplitude is 3 and there is no vertical translation or dilation parallel to the x-axis, the maximums would be at $\displaystyle \left (\frac{\pi}{2}\pm 2\pi n,\ 3\right )$ and the minimums would be at $\displaystyle \left (\frac{3\pi}{2}\pm 2\pi n,\ -3\right )$ (where $\displaystyle n$ is an integer).
As for the zeros, there would also be infinitely many, since there is no domain specified.
If the domain were $\displaystyle \left [0,2\pi \right]$ (i.e. one cycle of this particular function) the zeros would be at $\displaystyle 0, \pi$ and $\displaystyle 2\pi$
There's a good guide to basic trig graphs and functions here: http://www.purplemath.com/modules/triggrph.htm
since there is no specified domain, i will make this general:
zeros occur when sin3x=0,
so y=sin3x crosses 0 when x=....-60,0,60, 120.... => x=kpi/3 where k is an integer
to find max/mins we differentiate y=sin3x
you get y'=3cos3x
when y'=0 you get stationary => max/min
3cos3x=0, when x=....,-90, -30,30, 90... => (2k-1)pi/6
the mins and maxes alternate (max when k is odd, and min when k is even)
edit: sorry misread Q, thought it was sin3x, but it is actually 3sinx, but i will still leave this incase anyone needs it