3cos2x-7cosx+5=0 0<x<360

Question

thanks

Answer 1

Not clear if you meant:
3cos(2x)-7cosx+5=0 0<x<360
or
3cos^2 x-7cosx+5=0 0<x<360
big difference between the two.

if you want 3cos^2 x-7cosx+5=0 , then it is simply a quadratic in cosx
(compare to 3m^2 - 7m + 5 = 0)
and it has no real solutions

if you want 3cos(2x)-7cosx+5=0 then
3(2cos^2 x - 1) - 7cosx + 5 = 0
6cos^2 x - 3 - 7cosx + 5 = 0
6cos^2 x - 7x + 2 = 0 , another quadratic which factors
(3cosx-2)(2cosx-1) = 0
cosx = 2/3 or cosx = 1/2

if cosx = 2/3, x = appr 48.2° or 311.8° , using my calculator
if cosx = 1/2, x = 60° or 300°