How do I solve for x in this equation

3sin²x = cos²x, where 0 ≤ x < 2π
I really have no idea where to even start :/

3Sin^2 x - Cos^2x = 0

Now use this identity:

1-Sin^2x = Cos^2x

Substitute this value of Cos^2x in the above equation. You will get

3Sin^2x - ( 1 - Sin^2x) = 0

3Sin^2x - 1 + Sin^2x = 0

4 Sin^2x -1 = 0
Sin^2x = 1/4

Take under root
Sinx = + 1/2 or - 1/2


Use calculator to get inverse of Sin to get angle which will be 30.

30 degrees is ur basic angle

so the answer for +1/2 will be 30 degrees , 150 degress

for the -ve one it will be 210 and 330.

3sin²x = cos²x
3sin²x - cos²x = 0
3sin²x - (1 - sin²x) = 0
4sin²x - 1 = 0
(2sinx + 1)(2sinx - 1) = 0

sinx = ±½

x = arcsin(0.5), or
x = arcsin(-0.5)

x = π/6, 5π/6, 7π/6, 11π/6

There ya go!
alice