How do you solve #cos x = cot x# over the interval 0 to 2pi?

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Answer 1 · 2016-03-01T07:45:18

It is

#cos x = cot x=>cosx=(cosx)/sinx=#

First it should be (so the denominator can be defined)

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Hence we have that

#cos x = cot x=>cosx=(cosx)/sinx=>cosx*sinx=cosx=> cosx*(sinx-1)=0=> cosx=0 or sinx=1#

From #cosx=0=>x=pi/2 or x=3pi/2#

From #sinx=1=>x=pi/2#

Hence the solutions are #x=pi/2,x=3pi/2#