How can i find the domain of this?

Question

#f_x=x/(2cosx-1)#

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Answer 1 · 2018-01-19T15:29:44

See below.

#x/(2cosx-1)#

If the expression is as it's written, then there is no restriction on #x# Therefore the domain would be:

#{x in RR }#

If it is supposed to be:

#x/(2cos(x-1))#

Then the function is undefined if:

#2cos(x-1)=0#

#cos(x-1)=0#

#arccos(cos(x-1)=arccos(0)#

#x-1= pi/2 , (3pi)/2#

#x=(2+pi)/2 , (2+3pi)/2#

The function will also be undefined for multiples of these values.
Notice that if we add #pi# onto #(2+pi)/2# we get:

#pi+(2+pi)/2=(2+3pi)/2#

If we add #2pi# onto #(2+pi)/2# we get:

#(2+5pi)/2#

So we can add multiples of #pi# to #(2+pi)/2#

and all these values will give a zero denominator .

This is also true if we subtract multiples of #pi#

This is usually expressed as:

#npi+(2+pi)/2color(white)(888)# for #color(white)(88)n in ZZ# ( positive or negative integer )

So domain will be:

#{ x in RR : x != npi+(2+pi)/2}#