How do you solve: $tan^2 x - 1 = 0$ for solutions up to 2pi?

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Answer 1 · 2015-04-16T10:45:39

$x in {pi/4,(3pi)/4,(5pi)/4,(7pi)/4}$

If $tan^2(x) -1 =0$
then
$tan^2(x) = 1$
and
$tan(x) = +-1$

Within the domain $[0,2pi]$
this occurs at
$x = pi/4$ , $x=(3pi)/4$ , $x=(5pi)/4$ , and $x=(7pi)/4$
(from basic definition of $tan$ )

How do you solve: tan^2 x - 1 = 0tan^2 x - 1 = 0 for solutions up to 2pi?