solve algebraically, express the roots in exact form

2sinxcosx-1=0
I got:
2sinxcosx=1
sin^2x=1
what do I do next?

thanks in advance

Similar Questions
  1. Is this correct?(sinx - cosx)^2 = 1 - 2sinxcosx LS = sin^2x - 2sinxcosx + cos^2x = 1 - cos^2x - 2sinxcosx + cos^2x = 1 - cos^2x
    1. answers icon 0 answers
  2. Is this correct?(sinx - cosx)^2 = 1 - 2sinxcosx LS = sin^2x - 2sinxcosx + cos^2x = 1 - cos^2x - 2sinxcosx + cos^2x = 1 - cos^2x
    1. answers icon 1 answer
    1. answers icon 3 answers
  3. Solve for [0, 360)2sinxcosx + cosx =0 2sinxcosx = -cosx 2sinx = -cosx/cosx sinx = -1/2 {210, 330) Is this correct?
    1. answers icon 1 answer
more similar questions