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
(5y3)(y1) = 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(sinx1)^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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