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

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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 #

Explanation:

#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 #

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

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