Okay, I think I have gotten a little bit further, but I'm still stuck....

I have done the following:

2sin(x) -3 = 3cos(x)

2sin(x) -3 = 3 /sqrt [1-sin^2(x)]

[2sin(x) -3][2sin(x) -3] = 9 [1-sin^2(x)]

4sin^2(x) -6sin(x) - 6sin(x) + 9 = 9 - 9sin^2(x)

13sin^2(x) - 12sin(x) = 0

sin(x) [13sin(x) - 12] = 0

sin(x) = 0

Quadrants- I, II, III, and IV

Reference Angles= 0, 180, 360

x= 0, 180, 360

13sin(x) - 12 = 0

13sin(x) = 12

sin(x) = 12/13

x = 67.38

Quadrants- I and II

Reference angle= 67.38

0+67.38 = 67.38

180-67.38 = 112.62

x= 67.38 and 112.62

Therefore, x= 1, 67.38, 112.62, 180, and 360

But those values don't answer the question. The question asks to find what equation would have the intersection points of the graphs as its solutions?

So, I'm still lost....