Analytic Trigonometry. Sin(2u) - Sin(u) = 0
Sin(2u) - Sin(u) = 0 Sin(2u) = Sin(u) u=2u u=0
So what I know is that Sin(2u) - Sin(u) = 0 will convert into 2sin(u) * cos(u) - sin(u) = 0. I also know this is supposed to factor out but I do not know how to factor this
I don't think that you even need to expand the sin(2u), but this is just my opinion.
My pre-calculus book and the internet show me this. These are double angles
\[\Large\rm \left[2\sin u \cos u-\sin u\right]=0\]Factoring out sin u from each term gives you,\[\Large\rm \sin u\left[2\cos u-1\right]=0\]
Then apply your Zero-Factor Property:\[\Large\rm \sin u=0\]Which will produce the solutions that HelpAndNeed mentioned :) But it looks like we're getting another set of solutions from the other factor, yes?\[\Large\rm 2\cos u-1=0\]
I'm just not understanding how to factor [2sinucosu−sinu]=0. Total brainfart with this.
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