Note that as cos(x)cos(x) and sin(x)sin(x) will not both be 00 for the same xx, neither will be 00 for the given equation. Thus we may divide by cos^2(x)cos2(x) without introducing or losing any solutions.
cos^2(x) = 3sin^2(x)cos2(x)=3sin2(x)
=> (3sin^2(x))/cos^2(x) = cos^2(x)/cos^2(x)⇒3sin2(x)cos2(x)=cos2(x)cos2(x)
=> 3tan^2(x) = 1⇒3tan2(x)=1
=> tan^2(x) = 1/3⇒tan2(x)=13
=> tan(x) = +-1/sqrt(3) = +-sqrt(3)/3⇒tan(x)=±1√3=±√33
Now, through knowledge of well known angles or by examining the unit circle, we find that on the interval [0^@, 360^@)[0∘,360∘),
tan(x) = sqrt(3)/3 <=> x in {30^@, 210^@}tan(x)=√33⇔x∈{30∘,210∘}
and
tan(x) = -sqrt(3)/3 <=> x in {150^@, 330^@}tan(x)=−√33⇔x∈{150∘,330∘}
Putting those together, we get our solution set:
x in {30^@, 150^@, 210^@, 330^@}x∈{30∘,150∘,210∘,330∘}
