Show that 2cos^2(x) = 2 - 2sin^2(x) and hence solve 2cos^2(x) + 3sin(x) = 3 for 0<x<180

You can rearrange the equation to get 2cos^2(x)+2sin^2(x) = 2. This can be factorised to get cos^2(x) + sin^2(x) = 1 which is a known identity. We can use the fact that 2cos^2(x) = 2sin^2(x) - 2 and substitute this into the equation. This will give the equation 2sin^2(x) - 2 + 3sin(x) = 3. Rearranging you can get the equation into the form 2sin^2(x) + 3sin(x) - 5 = 0. From here we can substitute u = sin(x) to get the equation 2u^2 + 3u -5 = 0. You can then solve this quadratic using the quadratic formula to get that u = 1 or u = -5/2. Subbing sin(x) back in for u you get that sin(x) = 1 or sin(x) = -5/2. You can ignore the sin(x) = -5/2 as -1 < sin(x) < 1. This leaves that sin(x) = 1. Now to find the values of x between 0 < x < 180. You know that sin(90) = 1 or you can use the arcsin(x) function on your calculator to get the value 90 so you have found that x = 90.