Note that as cos(x) and sin(x) will not both be 0 for the same x, neither will be 0 for the given equation. Thus we may divide by cos^2(x) without introducing or losing any solutions.
cos^2(x) = 3sin^2(x)
=> (3sin^2(x))/cos^2(x) = cos^2(x)/cos^2(x)
=> 3tan^2(x) = 1
=> tan^2(x) = 1/3
=> tan(x) = +-1/sqrt(3) = +-sqrt(3)/3
Now, through knowledge of well known angles or by examining the unit circle, we find that on the interval [0^@, 360^@),
tan(x) = sqrt(3)/3 <=> x in {30^@, 210^@}
and
tan(x) = -sqrt(3)/3 <=> x in {150^@, 330^@}
Putting those together, we get our solution set:
x in {30^@, 150^@, 210^@, 330^@}
