have you come across
cos 2x = 2cos^2 x - 1 ?
let's sub it.
3cos 2x - 7cosx = 0
3(2cos^2 x - 1) - 7cosx = 0
6cos^2x - 7cosx - 3 = 0
(2cosx-3)(3cosx+1) = 0
cosx = 3/2 , which is not possible,
or
cosx = -1/3
so x must be in quadrants II or III
angle in standard position is 7.53 , then
x = 180-70.53 = 109.47°
or
x = 180+70.53 = 250.53
Solve 3cos2x - 7cosx = 0, when 0<=x<=360
And also, find the exact values for x when 0<=x<=360 if 3tan^2x=1
Thankyou!
3 answers
3tan^2 = 1
tan^2x = 1/3
tanx = ± 1/√3
x = 30°, 150°, 210° or 330°
tan^2x = 1/3
tanx = ± 1/√3
x = 30°, 150°, 210° or 330°
3cos2x+7cosx+3=0
1 answer