it's not x=0 or x=2/3 its sinx=0 or sinx=2/3
Hello the equation I want to solve for 0 <= x <= pi
This is what I have so far:
3(1-sin^2 x) + 2 sin x - 3 = 0
3 - 3sin^2 x + 2 sin x - 3 = 0
-3 sin^2 x + 2 sin x = 0
sin x (2 - 3sin x) = 0
roots: x = 0 or 2/3
Is this correct? But now what? Do I calculate cos(0)? and cos(2/3)?
No, that's not like that.
Usually though, you need to find the critical angle, and from then, you get the x that lie within the boundaries imposed to you, because in 2pi rad, you have two solutions for one sine ratio and in pi, you only have one solution most of the time. Here is one of those exceptions.
EDIT: I thought you did sin(2/3). Note that it's sin^-1(2/3) or arcsine(2/3)...