Question

Find all solutions if 0° ≤ θ < 360°

cos^2*3x−8cos3x+3=0

Answer 1

\[\cos ^{2}3x-8\cos 3x+3=0\]

Answer 2

set cos (3x) = t, you have t^2 -8t +3 = 0 . solve that quadratic which solutions are t1 = 7.6055551275 and t2= 0.39444487245. Reject t1 because cos of something cannot be bigger than 1. so you have cos (3x) = 0.394448725 therefore 3x = arccos (0.394448725) = 66.77 degree, so x = 22.26 degree.