How do you find all solutions of the equation in the interval $[0,2pi)$ given $sec^2x+6tanx+4=0$ ?

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Answer 1 · 2017-01-11T02:25:35

Apply the identity $sec^2x = 1 + tan^2x$ :

$1 + tan^2x + 6tanx + 4 = 0$

$tan^2x + 6tanx + 5 = 0$

Let $t = tanx$ .

$t^2 + 6t + 5 = 0$

$(t + 5)(t + 1) = 0$

$t = -5 and -1$

$tanx = -5 and tanx = -1$

$x = pi - arctan(5), 2pi - arctan(5), (3pi)/4, (7pi)/4$

Hopefully this helps!

How do you find all solutions of the equation in the interval [0,2pi)[0,2pi) given sec^2x+6tanx+4=0sec^2x+6tanx+4=0?