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Answer 1 · 2016-10-31T21:22:40

${sec}^{2} x + 2 tan x - 3 = 0$ $sec^2x + 2tanx - 3= 0$

Recall the pythagorean identity ${sec}^{2} θ = 1 + {tan}^{2} θ$ $sec^2theta = 1 + tan^2theta$ .

$1 + {tan}^{2} x + 2 tan x - 3 = 0$ $1 + tan^2x +2tanx - 3 = 0$

${tan}^{2} x + 2 tan x - 2 = 0$ $tan^2x + 2tanx - 2 = 0$

Let $t = tan x$ $t = tanx$

$t^{2} + 2 t - 2 = 0$ $t^2 + 2t - 2 = 0$

$t = \frac{- 2 \pm \sqrt{2^{2} - 4 \times 1 \times - 2}}{2 \times 1}$ $t = (-2 +- sqrt(2^2 - 4 xx 1 xx -2))/(2 xx 1)$

$t = \frac{- 2 \pm \sqrt{12}}{2}$ $t = (-2 +-sqrt(12))/2$

$t = \frac{- 2 \pm 2 \sqrt{3}}{2}$ $t = (-2 +- 2sqrt(3))/2$

$t = - 1 \pm \sqrt{3}$ $t = -1 +- sqrt(3)$

$tan x = - 1 \pm \sqrt{3}$ $tanx = -1 +- sqrt(3)$

$x = π n - arctan (\sqrt{3} - 1) and π n - arctan (1 + \sqrt{3})$ $x = pin - arctan(sqrt(3)-1) and pin - arctan(1 + sqrt(3))$

Hopefully this helps!