For problems like this, we should work inside out.
First, what values of #x# can go into the squareroot? We can't take a squareroot of a negative, so #x > 0#. Let's also realize that the squareroot can go from 0 to infinity.
Therefore, the number that goes into cosecant is within #(-infty, 0]#. Cosecant repeats for every #2 pi# so we know this will go through its entire range by #4pi^2 approx 39.5#. Therefore, the range of the cosecant (before multiplying by 3) is the usual: #(-infty, -1] cup [1,infty) #. The 3 gets multiplied through, meaning that in the end,
Domain: #[0, infty)#
Range: #(-infty, -3] cup [3, infty)#
So it crosses the #y# axis and exists in the 1st and 4th quadrants. Putting it into a plotting program confirms everything we just said:
graph{1/sin(0-x^(1/2)) [-5.55, 96.65, -20.24, 30.9]}
