Question

Pre-Calc Question: List all solutions between [ 0 , 2pi]. cos^2(x)sin(x) = sin(x)

Answer 1

I thought two of the solutions would be pi/2 and 3pi/2 since I simplified it to cos^2(x) = 0

Answer 2

I see a mistake I made. I subtracted sin(x) from both sides rather than dividing.

Answer 3

Let me try re-working it one second

Answer 4

I'm curious as to why you ended up going further with sin though. Wouldnt you just divide sin(x) from both sides and that would leave you with cos^2(x) = 1? Then go from there

Answer 5

You cannot divide by sinx. You can factor out a sinx, but that's going to lead to something more.

Answer 6

Move the sinx over, factor it out, and you have \[\cos ^{2}x \sin x=\sin x\] is this what you started out with?

Answer 7

Yes that is what I started with

Answer 8

ok, now do not divide by sinx, instead, move it over, set the entire thing to 0, and factor it...it will look like this:

Answer 9

sinx(cos^2x - 1) = 0 right?

Answer 10

\[\cos ^{2}x \sin x-\sin x=0\] which by facotring out a sinx becomes:
\[\sin x(\cos ^{2}x-1)=0\]

Answer 11

yes, now set each of those factors =0.

Answer 12

sinx=0 or cos^2x-1=0

Answer 13

cos^2x-1=0 factors into (cosx-1)(cosx+1)=0

Answer 14

can the cos^2x be interpreted as (cosx)^2? or is it cos(x^2)?

Answer 15

You can interpret cos^2x as (cosx)^2

Answer 16

Thank you very much for your help!

Answer 17

You are most welcome.