How many solutions does the equation 2sin^2(x) - 4sin(x) + cos^2(x) + 2 = 0 have in the domain 0<x<2pi

Firstly lets rearrange the equation to get rid of cos^2(x) to make it easier to solve by using the identity sin^2(x)+cos^2(x) = 1. Then we get 2sin^2(x) - 4sin(x) + 1 - sin^2(x) + 2 = sin^2(x) - 4sin(x) + 3 = 0
Now lets say y = sin(x) then our equation looks like this: y^2 - 4y + 3 = 0. We can factorise this into (y-1)(y-3)=0 which means y = 1,3. A common mistake now would be to say since y has 2 solutions sin(x) has 4 solutions in the specified domain. This is not true as it is important to look at what y really is. y is actually sin(x) which has a range of -1<sin(x)<1 for all x hence there are no solutions to the equation sin(x) = 3. The number of times sin(x) = 1 in our domain is once at x = pi/2 hence the number of solutions is in fact 1.