cos2x-3cosx=2(1+cosx)?

Question

Does anyone can tell me an answer? I cannot solve it!

Answer 1

2 cos ² x - 1 - 3 cos x = 2 + 2 cos x
2 cos ² x - 5 cos x - 3 = 0
(2 cos x + 1)(cos x - 3) = 0
cos x = - 1/2 is acceptable solution.
x = 120° , x = 240°

Answer 2

cos2x - 3 cosx = 2 + 2cosx
(2cos^2x -1) - 3cosx = 2 + 2cosx
2cos^2x -5cosx -3 = 0
(2cosx + 1)(cosx -3) = 0
cosx = -1/2
cosx = 3 but cosx is between 1 and -1 so this doesn't work

cosx = -1/2
x = cos^-1(-1/2)
x = 120 degrees and 240 degrees

Answer 3

Substituting cos2x= 2cos*2x-1
cos2x-3cosx=2(1+cosx)
2cos*2x-1-3cosx=2+2cosx
2cos*2x-5cosx-3=0
cosx=m
2m*2-5m-3=0
(m-3)(2m+1)=0
m=3 or -1/2
As cosx can not be more than 1, te solution cosx=3 does not exist here.
cosx= -1/2
cos(180-60)= - cos 60=-1/2
Therefore the possible answer is x= 120 degrees.

Answer 4

cos 2x - 3cosx = 2 + 2cosx
cos^2x - sen^2x - 3cosx = 2 + 2cosx
cos^2x - (1 - cos^2x) - 3cosx = 2 + 2cosx
cos^2x - 1 + cos^2x - 3cosx - 2 - 2cosx = 0
2cos^2x - 5cosx - 3 = 0
2cos^2x - 6cosx + cosx - 3 = 0
2cosx (cosx - 3) + cosx - 3 = 0
(cosx - 3) (2cosx + 1) = 0
cosx - 3 = 0
cosx = 3 is a non valid solution
or 2cosx + 1 = 0
2cosx = - 1
cosx = - 1/2
x = 120° + 2n(pi)

Answer 5

cos2x - 3cosx = 2 + 2cosx
cos2x = 2 + 5cosx

cos2x = 2cos^2x - 1

2cos^2x - 1 = 2 + 5cosx
2cos^2x = 3 + 5cosx
2cos^2x - 5cosx - 3 = 0

u = cosx
2u^2 - 5u - 3 = 0
(2u + 1) ( u - 3) = 0
u = 3, u = -1/2

3 = cosx
-1/2 = cosx

cos-1(3), cos(-1/2)

Answer 6

You can't solve it.... you are missing more properties if we are looking for x..