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Find all solutions in the interval [0, 2π). 4 sin2x - 4 sin x + 1 = 0 I'd like to know the identities used and how you came to know which one to use.
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(4 sin^2x) - 4 sin x + 1 = 0
youd use the quadratic equation ....
let u = sin(x)
4 sin^2 x - 4 sin x + 1 = 0 4 u^2 - 4u + 1 = 0 solve for u to determine sin(x)
Wow... I feel dumb. Thank you very much!
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youre welcome
Where do I go from (2u-1)^2 = 0?
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