helllpp in trig.?

Question

find the solution set of
4sin^2x -1 = 0 where x belongs to ] 0 , 180 [

Answer 1

4sin^2 x = 1
sin^2 x = 1/4 taking square root for both
sin x = +/1/2
x = sin^-1 1/2
x might lie in first or second or third or even fourth quadrant
cuz it has +/- sign and sinx is +ve in first and secon and negative in third and fourth
so we get x using calc = 30
then first quad = 30 second = 180 - 30 =150
third quad = 180+30 = 210 fourth quad = 360-30 = 330but you are given boundary condition that x belogs to ]0,180[
so first and secon quad only valid
solution = { 30 , 150 }
notice the other two answers are wrong
the first didn't put +/- gave you one angle
the second forgot the boundary condition x belongs ]0,180[

Answer 2

4sin ² x = 1
sin ² x = 1/4
sin x = 1/2 where x lies in 1st and 2nd quadrants
x = 30° , x = 150°

Answer 3

4sin^2 x = 1
sin^2 x = 1/4
sin x = 1/2
x = sin^-1 1/2
x = 30 and 150

Answer 4

sinx = +-1/2 so in the given interval x= 30 deg and x= 150deg. The sin is negative between 180 and 360 deg

Answer 5

4sin^2x -1 = 0
sin^2(x)=1/4
sinx=+-1/2
x=30 or 150 for [0,180]