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first n last are right but the second one is not
So the first one on 11?
\[\sin2x=2sinxcosx\] substitute this in re arrange to get the equation equal to 0 then take out a factor and that is your answer
yep try it @ohohaye
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Oh, ok
Thank you, I think I got it
Only the second one is wrong, The solution is as follows 5sin(2x)=3cosx => 5*2*sinx*cosx=3cosx =>10*sinx*cosx -3cosx=0 =>cosx(10sinx-3)=0
Oh, ok
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