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Mathematics 18 Online
OpenStudy (anonymous):

the most general solution which satisfies tan a=1, cos a=-1/sqrt()

OpenStudy (anonymous):

\[\tan \theta=1, \cos \theta=-1/\sqrt{2}\]

OpenStudy (anonymous):

\[\theta=\pi/4+\pi+2n \pi={5\pi \over 4}+2n \pi; n=0,1,2,..\]

OpenStudy (anonymous):

would you explain it a bit

OpenStudy (anonymous):

First, the angle at which tan is equal 1, and cos is 1/sqrt(2) is pi/4.

OpenStudy (anonymous):

Then we should look for the quadrant that has positive values for tan and negative values for cos, and that would be the third quadrant.

OpenStudy (anonymous):

So the correspondent angle for pi/4 in the third quadrant is pi/4 +pi.

OpenStudy (anonymous):

This angle will occur again after 2n(pi) for n=0,1,2,...

OpenStudy (anonymous):

Does that make sense?!

OpenStudy (anonymous):

\[2n \pi +3\pi/4\] is it rite?

OpenStudy (anonymous):

Yeah, if you have both cos and tan negatives.

OpenStudy (anonymous):

nd id i have jus the cos -ve it wud be 5pi/4...is that correct?

OpenStudy (anonymous):

*if

OpenStudy (anonymous):

Exactly.

OpenStudy (anonymous):

kool;-)

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