For what values of #x# does #tan(-2x) = cot(x)# ?

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Answer 1 · 2017-04-08T17:47:24

#x = ((2k+1)pi)/2" "# for any integer #k#.

Given:

#tan(-2x) = cot(x)#

That is:

#sin(-2x)/cos(-2x) = cos(x)/sin(x)#

Multiplying both sides by #cos(-2x)sin(x)#, this becomes:

#sin(x)sin(-2x) = cos(x)cos(-2x)#

Subtracting #sin(-2x)sin(x)# from both sides, we get:

#cos(x)cos(-2x)-sin(x)sin(-2x) = 0#

Compare the left hand side with the sum formula for #cos#:

#cos(alpha)cos(beta)-sin(alpha)sin(beta) = cos(alpha+beta)#

So with #alpha=x# and #beta=-2x# we get:

#cos(-x) = 0#

Note that #cos(-x) = cos(x)#

Hence:

#x = ((2k+1)pi)/2#

where #k# is any integer.

Here are the two functions #tan(-2x)# and #cot(x)# plotted together:

graph{(y+tan(2x))(y-cot(x)) = 0 [-10, 10, -5, 5]}