How do you solve $4tan^2x - 12tanx + 9 = 0$ from 0 to 2pi?

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Answer 1 · 2015-08-09T11:20:22

$x = 0.983 " or " x = pi+0.983 " for " 0 < x < 2pi$

$4 tan^2 x - 12 tan x + 9 = 0$
$(2tan x -3)^2 = 0$
$tan x = 3/2$
$x = tan^-1 (3/2) = 0.983$ (3sf)
$x = 0.983 " or " x = pi+0.983 " for " 0 < x < 2pi$

How do you solve 4tan^2x - 12tanx + 9 = 04tan^2x - 12tanx + 9 = 0 from 0 to 2pi?