Explanation:
sin^2(x)= 3 cos^2(x)
Subtract sin^2(x) on both sides
0=3cos^2(x)-sin^2(x)
By Pythagorean Identity sin^2(x)+cos^2(x)=1, we can replace cos^2(x) with 1-sin^2(x)
0=3(1-sin^2(x))-sin^2(x)
Distribute
0=3-3sin^2(x)-sin^2(x)
Combine like terms
0=3-4sin^2(x)
Add 4sin^2(x) on both sides
4sin^2(x)=3
Divide both sides by 4
sin^2(x)=3/4
Take the square root of both sides
sin(x)=+ or - sqrt(3/4)
sin(x)=+ or - sqrt(3)/sqrt(4)
sin(x)=+ or - sqrt(3)/2
This happens on the unit circle at
60 degrees
120 degrees
240 degrees
300 degrees
If we go around and around again forwards or backwards, we get the infinitely many other solutions including the one's already stated in the following solution sets
(60 +360n) degrees
(120 +360n)degrees
(240 +360n)degrees
(300 +360n)degrees
where n is an integer
x is 120 degrees or (120+360n) degrees in the second quadrant. I didn't know if you were just looking for solutions in the first nonnegative cycle of the unit circle or not.
240 degree one or the (240+360n) degrees terminates in the third.