How do you solve ${tan}^{2} (x) - | sec (x) | = 1$ $Tan^2(x)-|Sec(x)|=1$ ?

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Answer 1 · 2016-11-30T12:42:29

$x = 2 k π \pm \frac{π}{3}$ $x= 2kpi+-pi/3$ and

$x = 2 k π \pm \frac{2}{3} π, k = 0, \pm 1, \pm 2, \pm 3, . .$ $x= 2kpi+-2/3pi, k = 0, +-1, +-2, +-3, ..$

$| sec x | \geq 0$ $|sec x| >=0$ .

As ${tan}^{2} x = {sec}^{2} x - 1 = {| sec x |}^{2} - 1 .$ $tan^2x = sec^2x-1 =|sec x|^2-1.$

the given equation becomes a quadratic in $| sec x |$ $|sec x|$ .

${| sec x |}^{2} - | sec x | - 2 = (| sec x | - 2) (| x e c x | + 1) = 0$ $|sec x|^2-|sec x|-2=(|sec x|-2)(|xec x|+1)=0$ .

So, $| sec x | = 2 \to sec x = \pm 2$ $|sec x|=2 to sec x = +- 2$

${sec}^{- 1} (\pm 2) = \frac{π}{3} and \frac{2}{3} π$ $sec^(-1)(+-2)= pi/3 and 2/3pi$ .. And so,

$x = 2 k π \pm \frac{π}{3}$ $x= 2kpi+-pi/3$ and

$x = 2 k π \pm \frac{2}{3} π, k = 0, \pm 1, \pm 2, \pm 3, . .$ $x= 2kpi+-2/3pi, k = 0, +-1, +-2, +-3, ..$

How do you solve Tan^2(x)-|Sec(x)|=1 tan2(x)−|sec(x)|=1 Tan^2(x)-|Sec(x)|=1 ?