Full solution of "2cos^2x = 1 + sinx"

I'm assuming you would move everything over to the left side to start with, then use two variables. As in, replace cosx for x and sinx for y turning the equation into "2x^2 - y - 1". Don't know where to go from there, I'm not sure how to factor with two variables...

First, cos^2(x) = 1 - sin^2(x)
so the first step is to have only one variable (sin(x))
so:
2cos^2(x) = 1 + sin(x)
==> 2(1 - sin^2(x)) = 1 +sin(x)
==> 1 + sin(x) - 2 + 2sin^2(x) = 0
==> 2sin^2(x) + sin(x) - 1 = 0
==> (2sin(x) -1)(sin(x)+ 1) = 0
so: sin(x) = 1/2 or sin(x) = -1
sin(x) = 1/2
==> x = pi/6 (or 30 degrees)
or 5pi/6 (150 degrees) plus multiples of 2pi(360 degrees)

sin(x) = -1
==> x = -pi/2 = 3pi/2 (270 degrees) plus multiples of 2pi (360 degrees)

use the trigonometric identity that cos^2(x) = 1-sin^2(x) to get everything in terms of sin(x) and solve......