Question

solve:
sin^2x=sinxcosx

Answer 1

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
\[x= 2n \pi + \pi/4\]

Answer 2

The answer is
x=0, pi over 4, pi, 5pi over 4.

and how would I get 5pi over 4?

Thanks.

Answer 3

yeah sorry,it'll also include
\[n \pi+ \pi/4\]

Answer 4

so you'll get, 0 , pi/4 and 5pi/4

Answer 5

I think I got it.
Thanks.

Answer 6

welcome