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anonymous
 5 years ago
solve:
sin^2x=sinxcosx
anonymous
 5 years ago
solve: sin^2x=sinxcosx

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ash2326
 5 years ago
Best ResponseYou've already chosen the best response.4sin^2x =sin x cosx sin^2 xsinx cos x=0 sinx (sin xcos x)= either sin x=0 or sin x = cos x sin x =0 so \[x= n \pi\] sin x = cos x \[x= 2n \pi + \pi/4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The answer is x=0, pi over 4, pi, 5pi over 4. and how would I get 5pi over 4? Thanks.

ash2326
 5 years ago
Best ResponseYou've already chosen the best response.4yeah sorry,it'll also include \[n \pi+ \pi/4\]

ash2326
 5 years ago
Best ResponseYou've already chosen the best response.4so you'll get, 0 , pi/4 and 5pi/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I got it. Thanks.
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