Question
Im having problems solving this equation:
2sin(x) = cos(x) + 2
I keep getting an imaginary number..:/
2sin(x) = cos(x) + 2
I keep getting an imaginary number..:/
Answers
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
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
jess
im a bit confused as to where the 8sin(x) came from
drwls
It somes from squaring 2sin x - 2.
[2(sin x -1)]^2 = 4(sinx-1)^2
= 4 (sin^2 - 2x +1)= ...
[2(sin x -1)]^2 = 4(sinx-1)^2
= 4 (sin^2 - 2x +1)= ...
jess
ah, thank you, i don't know why i was squaring it differently (wrong way)
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
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