Mathematics
7 Online
OpenStudy (anonymous):
Solve for x: sin^2x+cos2x-cosx=0
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OpenStudy (anonymous):
i just tried yours and it's hard lol
OpenStudy (anonymous):
ik haha its freaking impossible
stupid summer assignment :/
hartnn (hartnn):
do u know the formula for sin ^2 x and cos 2x in terms of cos x and/or cos ^2 x ?
and its easy :)
OpenStudy (anonymous):
lol im trying to solve it right now
OpenStudy (anonymous):
\[\cos^2θ −\sin^2θ\] do we have to use that?
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OpenStudy (anonymous):
@hartnn
OpenStudy (anonymous):
The answer is \[x=2\pi n\]
hartnn (hartnn):
yes,u can use that,but that still contains sin^2 theta,can u convert that also in cos ^2 theta?
OpenStudy (anonymous):
no you can't it'll cancel though.. hold on let me do it on paper and then ill scan it.
OpenStudy (anonymous):
\[n\]
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OpenStudy (anonymous):
convert sin^2x into cos^2x
OpenStudy (anonymous):
\[n\]
OpenStudy (anonymous):
is this in terms of degrees or radians?
OpenStudy (anonymous):
radians
OpenStudy (anonymous):
okay im gonna scan my work now, i think it's right.
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OpenStudy (anonymous):
is there a range?
OpenStudy (anonymous):
\[0 \le x <2\]
OpenStudy (anonymous):
thats the domain
OpenStudy (anonymous):
2pi^
OpenStudy (anonymous):
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OpenStudy (anonymous):
im not sure if it's right.
OpenStudy (anonymous):
i thinkit is right
OpenStudy (anonymous):
i think so i too lol i don't know.
hartnn (hartnn):
yes its right,but 2pi should also be the answer
0,pi,2pi
OpenStudy (anonymous):
right
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OpenStudy (anonymous):
no there is no 2 pi
OpenStudy (anonymous):
nvrmd ur right my bad :P
OpenStudy (anonymous):
but r u sure cos2x= cos^2x-sin^2x
hartnn (hartnn):
ohh x<2pi
ok then 2pi will not be answer
yes that formula is correct
OpenStudy (anonymous):
ok is that like an identity or something?
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hartnn (hartnn):
yes,standard formula,which u can prove
OpenStudy (anonymous):
ok thx! do u have idea what its called?
OpenStudy (anonymous):
or is it just called standard formula? lol