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cos2x+3sinx-2=0
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\[\cos2x+3sinx-2=0 \] OR \[\cos^2x+3sinx-2=0 \] ?
first one
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
x=30,90,150
you can ask where you don't understand
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so factoring I factor a sin out
yes \[−2\sin^2x+3sinx−1=0\] \[(−sinx+1)(2sinx−1)=0\]
ok got it.Im having so much difficulty understanding these identities. Thanks
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