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Mathematics 6 Online
OpenStudy (anonymous):

Factor the algebraic expression below in terms of a single trigonometric function. cos x - sin ^2x - 1

OpenStudy (anonymous):

\[\cos x - \sin ^2x - 1 \]

OpenStudy (anonymous):

How do I do this?

OpenStudy (anonymous):

@RadEn

OpenStudy (anonymous):

cos x-(1-cos ^2 x)-1 =cos ^2x+cos x-2 =cos ^2x+2cos x-cos x-2 now you can make factors.

OpenStudy (anonymous):

Did that come out to \[\cos^2 x + 2\cos x-\cos x - 2\] @surjithayer ?

OpenStudy (anonymous):

\[ \cos^2 x + 2\cos x-\cos x - 2=\\ (\cos^2 x + 2\cos x)-(\cos x +2 )= \cos (x) (\cos(x) +2)- (\cos x +2 ) \] Can you finish it now?

OpenStudy (anonymous):

I got cos x(cos x +2) -1(cos x +2) Does that work @eliassaab

OpenStudy (anonymous):

correct,proceed further.

OpenStudy (anonymous):

I don't know what to do after that. Do you mean solve the zeros?

OpenStudy (anonymous):

=(cos x+2)(cos x-1)

OpenStudy (anonymous):

Oh, and that's it?

OpenStudy (anonymous):

I guess so. Thanks guys! God bless you!

OpenStudy (anonymous):

np

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