2sin^2x-sinx=0 - Math Central

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Question from Jose, a student:

2sin^2x-sinx=0
and i know the answer is this..
2sin²(x) - sin(x) = 0
sin(x)(2sin(x)-1) = 0

sin(x) = 0
x = 0

2sin(x) - 1 = 0
sin(x) = 1/2
x = π/6

x = {0, π/6}

im just having trouble figuring out how it went from the original equation to sin(x)2sin(x)-1=0?

Jose,

sin²(x) is sin(x) × sin(x) so 2sin²(x) - sin(x) = 2 sin(x) × sin(x) - sin(x) and this expression has a common factor of sin(x) so

2sin²(x) - sin(x) = 2 sin(x) × sin(x) - sin(x) = sin(x) [2 sin(x) - 1]

Harley

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