Posted by **jess** on Tuesday, December 18, 2007 at 8:02pm.

Im having problems solving this equation:

2sin(x) = cos(x) + 2

I keep getting an imaginary number..:/

- math (trig) -
**drwls**, Tuesday, December 18, 2007 at 8:21pm
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

(5y-3)(y-1) = 0

sin x = 1 or 3/5

x = pi/2 (90 degrees) is one solution. There are others as well

- math (trig) -
**drwls**, Tuesday, December 18, 2007 at 8:33pm
It somes from squaring 2sin x - 2.

[2(sin x -1)]^2 = 4(sinx-1)^2

= 4 (sin^2 - 2x +1)= ...

- math (trig) -
**Reiny**, Tuesday, December 18, 2007 at 9:15pm
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

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