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Mathematics 7 Online
OpenStudy (anonymous):

Solve for x: sin^2x+cos2x-cosx=0

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?

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):

\[n\]

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.

OpenStudy (anonymous):

is there a range?

OpenStudy (anonymous):

\[0 \le x <2\]

OpenStudy (anonymous):

thats the domain

OpenStudy (anonymous):

2pi^

OpenStudy (anonymous):

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

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?

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

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