How do you solve #cos^-1(tan x) = pi#?

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Answer 1 · 2018-03-24T03:45:11

#x = (4n+3)pi/4, qquad n in ZZ#

#cos^-1(tan x) = pi implies tan x = cos pi = -1#

So #x = tan^-1(-1) #

Now, in the interval #[0,2pi)#, there are two angles that satisfy #tanx = -1# - these are #x = (3pi)/4# and #(7pi)/4#.

The general solution is

#(3pi)/4 + 2npi# or #(7pi)/4+2n pi#, #qquad n in ZZ#

The two can be combined in a single expression

#x = (4n+3)pi/4, qquad n in ZZ#