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

$x = 0.983 \text{ or " x = pi+0.983 " for } 0 < x < 2 \pi$
$4 {\tan}^{2} x - 12 \tan x + 9 = 0$
${\left(2 \tan x - 3\right)}^{2} = 0$
$\tan x = \frac{3}{2}$
$x = {\tan}^{-} 1 \left(\frac{3}{2}\right) = 0.983$ (3sf)
$x = 0.983 \text{ or " x = pi+0.983 " for } 0 < x < 2 \pi$