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solve sinx + 2sinxcosx = 0, [0,2pi)
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Can you factor something out, so you have a product that is =0?
And if you can, then since AB=0, you know that A=0 or B=0. That will give you 2 simpler equations to solve. :)
?
sinx(1 + 2cosx) =0
is it sinx=0, cosx=-1/2
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Those are your equivalent compound equations, yes. :) Now you need to solve each of those. You should get 2 solutions for each one, so 4 total solutions.
pi, 2pi, 2pi/3, and 4pi/3 ?
they want the solutions in the interval [0,2pi) the ) means up to but not including 2pi. Use 0 instead
thanks
Yeah. You're welcome.
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