How do I solve for x in degrees given sin 2x - 1 = 0?

Replace #2x# with #u#

#sin(u)-1=0#
#sin(u)=1#

#u=pi/2+2npi# where n is any integer.

We know that #pi# gives us #1# for #sin(u)#; however, since we're not told that we're working on an interval, we must account for all other possible solutions (there are infinitely many values that give #1# that repeat on a period of #2pi# , such as #+-(5pi)/2, +-(9pi)/2, +-(13pi)/2#) by adding an integer multiplied by #2pi# to our solution.

Switch back to #2x#:

#2x=pi/2+2npi#

Simplify so that we only have an #x# on the lefthand side :
#x=(pi/2+2npi)/2#

#x=pi/4+npi#

To convert radians by degrees, multiply by #180/pi#

#cancel(pi)/4 * 180/cancel(pi) = 180/4=45°#

#(ncancel(pi))*180/cancel(pi)=(180n)# where n is any integer.

Add these two back together:

#x=45°+180°n#

I am not sure whether you mean #sin(2x - 1)# or #sin (2x) - 1#, so I'm going to do both.

1) #sin(2x - 1) = 0#

However, we know that #sin(npi) = 0#, for any integer #n#.

This implies #2x-1 = npi#, therefore

#x = (npi+1)/2#, any #n -> "Z"#

and in degrees :

#x = (180°n + 1)/2#, any #n -> "Z"#.

2) #sin(2x) - 1 = 0#

#sin(2x) = 1#

We know that #sin(((2n-1)pi)/2) = 1#, again, for any integer #n#.

This implies #2x = ((2n-1)pi)/2#, therefore

#x = ((2n-1)pi)/4#, any #n -> "Z"#.

and in degrees :

#x = ((2n-1)180°)/4#, any #n-> "Z"#.