Question #d211c

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Answer 1 · 2017-12-09T03:04:45

#x inRR | x != pi/2 2pik#

Assuming, #tan(x)/3-5cos(x)#, first break tangent down in to it's components

#tan(x)/3-5cos(x) hArr (sin(x)/cos(x))/3 - 5cos(x)#

Simplifying we get

#sin(x)/(3cos(x)) - 5cos(x)#

The domain of the function is only going to be restricted by #3cos(x) = 0# so let's set that equal to zero and solve for x

#3cos(x) = 0 rarr cos(x) = 0#

#cos(x)# is zero at #pi/2# the domain is thus #x inRR | x != pi/2 2pik#