How do you solve sin2θ−cosθ=0 between -pi/2 less than or equal to θ less than or equal to 3pi/2?

Answer 1

sin2θ−cosθ=0
2sinØcosØ - cosØ = 0
cosØ(2sinØ - 1)

cosØ = 0 or sinØ = 1/2

from cosØ = 0 , where does the cosine curve cut the x-axis in your domain?
http://www.wolframalpha.com/input/?i=y+%3D+cosx

where is sinØ = 1/2 in your domain?
http://www.wolframalpha.com/input/?i=y+%3D+sinx+,+y+%3D+1%2F2

Answer 2

Thank you so much.