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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