How do you solve $tan^2x+4=2sec^2x+tanx$ in the interval [0,360]?

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Answer 1 · 2016-11-05T01:33:44

Use the identity $sec^2theta = 1 + tan^2theta$ .

$tan^2x + 4 = 2(1 + tan^2x) + tanx$

$tan^2x + 4 = 2 + 2tan^2x + tanx$

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

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

$0 = tanx(tanx + 2) - 1(tanx + 2)$

$0 = (tanx - 1)(tanx + 2)$

$tanx = -2$ and $tanx = 1$

$x = 180˚ - arctan(2)$ , $360˚- arctan(2), 45˚, 225˚$

Hopefully this helps!

How do you solve tan^2x+4=2sec^2x+tanxtan^2x+4=2sec^2x+tanx in the interval [0,360]?