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!