math (trig)
posted by jess .
Im having problems solving this equation:
2sin(x) = cos(x) + 2
I keep getting an imaginary number..:/

2 sin x 2 = cos x = sqrt(1  sin^2 x)
4 sin^2 x  8 sin x + 4 = 1  sin^2 x
5 sin^2 x  8 sin x + 3 = 0
Let sin x = y
5y^2 8y +3 = 0
(5y3)(y1) = 0
sin x = 1 or 3/5
x = pi/2 (90 degrees) is one solution. There are others as well 
im a bit confused as to where the 8sin(x) came from

It somes from squaring 2sin x  2.
[2(sin x 1)]^2 = 4(sinx1)^2
= 4 (sin^2  2x +1)= ... 
ah, thank you, i don't know why i was squaring it differently (wrong way)

Jess, don't forget that after you square an equation all answers you obtain must be verified.
for example, the second part to drwls solution was
sinx = 3/5
which produces answers of 36.87º and 143.13º
when these are checked in the original equation, x= 143.13 works but x = 36.87 does not
so x = 90º or x = 143.13º