Asked by jess

Im having problems solving this equation:

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

I keep getting an imaginary number..:/

Answers

Answered by drwls
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
Answered by jess
im a bit confused as to where the 8sin(x) came from
Answered by drwls
It somes from squaring 2sin x - 2.
[2(sin x -1)]^2 = 4(sinx-1)^2
= 4 (sin^2 - 2x +1)= ...

Answered by jess
ah, thank you, i don't know why i was squaring it differently (wrong way)
Answered by Reiny
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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