Graphing a trig function

[cruxxfay]

The question is as follow:
f(x) = (2+3sinx)/(4+3cosx)

I know that the first thing you do is to try simplifying this down to 1 trig ratio, but well, I use cos(x) = sqrt[1-sin^2(x)], which got me nowhere.
I have looked at the answer and the shape appears pretty weird to me so I have no idea where to start.

Thanks in advance

[jg.allinsymbols]

You might try looking for what is simple to do, first. Plug well know values in for x, and find the result of f(x). Plot each point. Make a table for x, 3sin(x), 2+3sin(x), 3cos(x), 4+cos(x). Try including a the domain from -2pi to +2pi.

If you'll try the rational expression as function on graphing calculator, you will see it is a sawtooth shape, raised above the horizontal axis.

[cruxxfay]

So you mean I just plug in pi/6, pi/4, pi/3 etc...? Because I was kind of looking for a solution instead of using the substitution method

[jg.allinsymbols]

Yes. I mean substitute known values for sine and cosine and compute the values for f(x), and then plot the points. Hopefully some other member reading this may have better qualitative knowledge about graphing more complicated Trigonometric composed functions and knows how to identify patterns in them. For now, creating a table of values from which to plot points, or using a graphing calculator are your best ways to find the graph.

[jg.allinsymbols]

Here is another thought:

Restrict the calculated set of values to a partcular domain, and USE THAT as the pattern; just offset the next set of plotted points according to this pattern. Sine and cosine are still periodic. Maybe, say plot points for 0 to 2*pi, and then the next set of plotted points will just be offset horizontally from the first set of points.