just need some help getting started

1.) sin2x= -1/2

2.) 4cos^2 x-8cosx+1=0

3.) sin2x= -sinx

greatly appreciate it(Nod)

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- Nov 28th 2009, 04:47 PMzukywich08inverse trigonometric functions
just need some help getting started

1.) sin2x= -1/2

2.) 4cos^2 x-8cosx+1=0

3.) sin2x= -sinx

greatly appreciate it(Nod) - Nov 28th 2009, 05:10 PMskeeter
- Nov 28th 2009, 05:31 PMzukywich08quick question
for the third problem how would you factor it? do you take sinx out of it?

- Nov 28th 2009, 05:34 PMskeeter
- Nov 28th 2009, 05:39 PMzukywich08
i got sinx(sinxcosx+1)=0

but i dont know what to do with the cosine? - Nov 28th 2009, 05:41 PMskeeter
- Nov 28th 2009, 05:56 PMzukywich08
alright thanks for the help man it came in handy