Maths  trig
posted by Rose on .
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!

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
(2cosx3)(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 = 18070.53 = 109.47°
or
x = 180+70.53 = 250.53 
3tan^2 = 1
tan^2x = 1/3
tanx = ± 1/√3
x = 30°, 150°, 210° or 330° 
3cos2x+7cosx+3=0