Here's the question you clicked on:
Hannah_Ahn
solve: sin^2x=sinxcosx
sin^2x =sin x cosx sin^2 x-sinx cos x=0 sinx (sin x-cos x)= either sin x=0 or sin x = cos x sin x =0 so \[x= n \pi\] sin x = cos x \[x= 2n \pi + \pi/4\]
The answer is x=0, pi over 4, pi, 5pi over 4. and how would I get 5pi over 4? Thanks.
yeah sorry,it'll also include \[n \pi+ \pi/4\]
so you'll get, 0 , pi/4 and 5pi/4
I think I got it. Thanks.