How do you solve for x in: ${tan}^{2} x - 3 tan x - 4 = 0$ $tan^2 x - 3 tan x - 4 = 0$ ?

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Answer 1 · 2015-04-15T12:36:39

Let $k = tan (x)$ $k=tan(x)$
then
${tan}^{2} (x) - 3 tan (x) - 4 = 0$ $tan^2(x) -3tan(x)-4 = 0$
becomes
$k^{2} - 3 k - 4 = 0$ $k^2-3k-4=0$

$(k - 4) (k + 1) = 0$ $(k-4)(k+1) = 0$

So we are looking for
$k = 4$ $k=4$ or $k = - 1$ $k=-1$
that is, for
$tan (x) = 4$ $tan(x) = 4$ or $tan (x) = - 1$ $tan(x)=-1$

$x = arctan (4)$ $x = arctan(4)$
or
$x = arctan (- 1) = - \frac{π}{4} + n (π)$ $x= arctan(-1) = -pi/4 +n(pi)$ for all $n epsilon ZZ$

How do you solve for x in: tan^2 x - 3 tan x - 4 = 0tan2x−3tanx−4=0tan^2 x - 3 tan x - 4 = 0?