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Find all solutions in the interval [0, 2π). 2 sin2x = sin x
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sin^2x*
Subtract sin x from both sides and factor out sin x. Then set each factor equal to 0 and solve.
Okay, thanks. So you factor and set it to 0?
yes. Set each factor to 0
I'm sorry, I had a blank moment, how do you factor? Once again, so sorry about the stupid question.
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\[2 \sin^{2} x = \sin x\] \[2\sin^{2} x - \sin x = 0\] \[\sin x (2 \sin x - 1) = 0\]
Okay, I did it correctly, thought I did it wrong. :P
Thanks to the both of you.
yw
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