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Mathematics 12 Online
OpenStudy (anonymous):

cos2x+3sinx-2=0

sam (.sam.):

\[\cos2x+3sinx-2=0 \] OR \[\cos^2x+3sinx-2=0 \] ?

OpenStudy (anonymous):

first one

sam (.sam.):

You have to make it like a quadratic equation, Use the identity \[\cos2x=1-2\sin^2x\] ---------------------------- From equation, \[\cos2x+3sinx-2=0 \] \[1-2sin^2x+3sinx-2=0 \] \[-2sin^2x+3sinx-1=0 \] Factor, then you'll get sinx=1, sinx=1/2

sam (.sam.):

x=30,90,150

sam (.sam.):

you can ask where you don't understand

OpenStudy (anonymous):

so factoring I factor a sin out

sam (.sam.):

yes \[−2\sin^2x+3sinx−1=0\] \[(−sinx+1)(2sinx−1)=0\]

OpenStudy (anonymous):

ok got it.Im having so much difficulty understanding these identities. Thanks

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