HELP!!!!! (1 Viewer)

[MartinsPlace]

please solve these two equations for me. i keep getting it wrong.

2cosx = 1+secx

and cos2x=4sin^2x - 14cos^2x


thanks

 

[Leffife]

Okay okay

1)
2cos x = 1 + sec x
2cos x = 1 + 1/cos x
2cos²x = cos x + 1
2cos²x - cos x - 1 = 0
(2cosx + 1) (cosx - 1) = 0
cos x = 1 , -1/2
Therefore, x = 0, 2π/3 , 4π/3 , 2π

2)
cos 2x = 4 sin² x - 14 cos² x
= 4[(1-cos 2x)/2] - 14[(1+cos 2x)/2]
10cos2x = - 5
cos 2x = -1/2
2x = 2π/3 , 4π/3 , 8π/3 , 10π/3
Therefore, x = π/3 , 2π/3 , 4π/3 , 5π/3

EDIT:
I hope you know how I got (1 - cos 2x) and (1 + cos 2x). Also I skipped some steps after "4[(1-cos 2x)/2] - 14[(1+cos 2x)/2]".

If you need the full thing then after that do.
cos 2x=2(1 - cos 2x) - 7(1 + cos 2x)
cos 2x= 2 - 2cos 2x - 7 - 7cos 2x
cos 2x= -9cos 2x - 5
10cos 2x = -5
Then it goes cos 2x = -1/2 etc.