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^@}#
