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

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

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