How do you solve 4tan^2x - 12tanx + 9 = 04tan2x−12tanx+9=0 from 0 to 2pi?

1 Answer
dani83
Aug 9, 2015

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

Explanation:

4 tan^2 x - 12 tan x + 9 = 04tan2x−12tanx+9=0
(2tan x -3)^2 = 0(2tanx−3)2=0
tan x = 3/2tanx=32
x = tan^-1 (3/2) = 0.983x=tan−1(32)=0.983 (3sf)
x = 0.983 " or " x = pi+0.983 " for " 0 < x < 2pi x=0.983 or x=π+0.983 for 0<x<2π

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