Finding all solutions of a trig equation

[jtw2e2]

Find all the solutions of the following trig equation with the condition 0 ? ? < 2?

2sin[sup:3idrpkvr]2[/sup:3idrpkvr]x-1=0

My work which is most likely incorrect and useless:

2sin[sup:3idrpkvr]2[/sup:3idrpkvr]x=1
sin[sup:3idrpkvr]2[/sup:3idrpkvr]x= 1/2
sinx = ±?(1/2)
sinx = ±(1/4)

This value doesn't really help me, especially since it's not one of our memorized angles on the unit circle. Help please

 

[wjm11]

Find all the solutions of the following trig equation with the condition 0 ? ? < 2?

2sin2x-1=0

My work which is most likely incorrect and useless:

2sin2x=1
sin2x= 1/2
sinx = ±?(1/2)
sinx = ±(1/4)

Good, until the last step. (1/2)^.5 does NOT equal 1/4.

sinx = ±?(1/2) = +/- ((2)^.5) /2

That should help a little...

 

[Deleted member 4993]

Another way:

\(\displaystyle 1 - 2sin^2(x) = cos(2x)\)

 

[jtw2e2]

Thanks to wjm11 for pointing that out about my error on the ?(1/2).

sinx = ±?2/2

That did help a lot and I got the correct values.