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Mathematics 9 Online
OpenStudy (anonymous):

3cos^2(x)-sin^2(x)=0

OpenStudy (anonymous):

have to find the soultions on the interval [0, 2pi)

OpenStudy (anonymous):

Use the identity cos^2(x)=1-sin^2(x) and you will have a problem in terms of sin^2(x)

OpenStudy (anonymous):

thank you!

OpenStudy (anonymous):

but that doesn't work out.. unless im doing it wrong..?

OpenStudy (anonymous):

It goes: 3(1-sin^2(x))-sin^2(x)=0 3-3sin^2(x)-sin^2(x)=0 So -4sin^2(x)=-3

OpenStudy (anonymous):

okay.. so then sin(x)= sq root (3/4) but sin never equals the sq root (3/4) on the unit circle?

OpenStudy (anonymous):

never mind. that would be the same as sq root (3) / 2 duhhh! hah i figured it out. thank you!!

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