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Mathematics 14 Online
OpenStudy (anonymous):

solve this trig equation:

OpenStudy (anonymous):

\[4\sin ^{2}xcos ^{2}x=1-\cos ^{2x}\]

OpenStudy (anonymous):

write cos2x as 1-2sin^2x

OpenStudy (anonymous):

or is it cos^2x?

OpenStudy (anonymous):

so it becomes: \[4\sin ^{2}(1-2\sin ^{2}x)=1-(1-2\sin ^{2}x)\]

OpenStudy (anonymous):

cos^2

OpenStudy (anonymous):

ok forget then what i've written

OpenStudy (anonymous):

so..u cant use that

OpenStudy (anonymous):

yep, i need to use 1-sin^2x

OpenStudy (anonymous):

cos^2 x ==== 1- sin^2 x

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

so we get: \[4\sin^2(x) (1-\sin^2(x))=1-1-\sin^2(x)\]

OpenStudy (anonymous):

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