Find all solutions if 0° ≤ θ < 360°: cos^2(3θ)-7cos(3θ)+3=0

Find all solutions if 0° ≤ θ < 360°. When necessary, round your answers to the nearest tenth of a degree.

cos^2(3θ)-7cos(3θ)+3=0

Answers should be in this format, six answers.
θ= ___ °
θ= ___ °
θ= ___ °
θ= ___ °
θ= ___ °
θ= ___ °

cos²(3θ) − 7cos(3θ) + 3 = 0

Solve using quadratic formula (where variable is cos(3θ) instead of x)

cos(3θ) = (7 ± √(49−12)) / 2
cos(3θ) = (7 ± √37) / 2

cos(3θ) = (7+√37)/2 = 6.541381265 > 1 -----> no solutions

cos(3θ) = (7−√37)/2 = 0.458618735
3θ = arccos((7−√37)/2) + 360n, −arccos((7−√37)/2) + 360m
θ = 1/3 arccos((7−√37)/2) + 120n
θ = −1/3 arccos((7−√37)/2) + 120m

For n = 0 to 2, and m = 1 to 3, we get:
θ = 20.9°, 99.1°, 140.9°, 219.1°, 260.9°, 339.1°

0° ≤ θ < 360°: cos^2(3θ)-7cos(3θ)+3 = 0?

cos(3θ) = [7 +/- sqrt(7^2 - 4*3)] / 2,

cos(3θ) = [7 +/- 6.1] / 2
SO,
cos(3θ) = 6.55, ===> NOT Possible,
Alsso,
cos(3θ) = 0.45, ===> 3θ = arccos(0.45)
3θ = 63.3°, and, 3θ = -63.3°
Hence,
θ = 21.1° >==========================< ANSWER
AND
θ = -21.1° >========================< ANSWER

Holy Nargles... I dont know how to do that...