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OpenStudy (anonymous):
solve abs(tanx)=1 ?
would it be tan(x)=1 tan(x)=-1
and then the pi values?
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OpenStudy (freckles):
It is like solving tan^2(x)=1
which give the two equations tan(x)=1 or tan(x)=-1
OpenStudy (freckles):
And yeah |f(x)|=f(x) or -f(x)
depending our values of x in which f is positive or negative
OpenStudy (anonymous):
there is no interval asked in the question so would it be like x = 3pi/4 +-pik?
OpenStudy (freckles):
Well that is one equation...there should be another
OpenStudy (freckles):
wait that should be fine
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OpenStudy (anonymous):
yeah the one for tanx = 1
OpenStudy (anonymous):
and for tanx = 1
OpenStudy (anonymous):
i meant -1
OpenStudy (freckles):
yeah we need the one for tan(x)=1 also
OpenStudy (anonymous):
then i think i got it thank you :D
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OpenStudy (freckles):
I would have said x=pi/4+npi , -pi/4+npi
OpenStudy (anonymous):
well i got 3pi/4 +-pik and pi/4+-pik
OpenStudy (freckles):
same thing
OpenStudy (freckles):
my presents any integers
(not just positive)
you don't need +/- unless you want your k's to be nonnegative integers
OpenStudy (freckles):
I was talking about my n represents any integer*
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OpenStudy (anonymous):
:O i didnt know that
OpenStudy (anonymous):
ty
OpenStudy (freckles):
But you can leave it that way
It is fine
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