Asked by Anonymous
How do you solve sin2θ−cosθ=0 between -pi/2 less than or equal to θ less than or equal to 3pi/2?
Answers
Answered by
Reiny
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
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
Answered by
Anonymous
Thank you so much.
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