How to solve this equation? The topic is Radian Measure of Angle size.

Question

Solve the equation for 0 ≤ x ≤ 2π.

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Answer 1 · 2018-07-25T06:08:55

#x=pi/3, (4pi)/3,(3pi)/4, (7pi)/4#

#tan^2x+tanx=sqrt3 tanx+sqrt3#

#tan^2x+tanx-sqrt3tanx-sqrt3=0#

#tanx(tanx+1)-sqrt3(tanx+1)=0#

#(tanx-sqrt3)(tanx+1)=0#

#tanx-sqrt3=0# or #tanx+1=0#

#tanx-sqrt3=0#
#tanx=sqrt3#
#x=pi/3, pi+pi/3# --> #tan x# is positive in the first and third quadrant
#x=pi/3, (4pi)/3#

#tanx+1=0#
#tanx=-1#
#x=pi-pi/4, 2pi-pi/4# --> #tanx# is negative in the second and fourth quadrant
#x=(3pi)/4, (7pi)/4#

Answer 2 · 2018-07-25T06:12:17

#pi/3, (3 pi)/4, (4 pi)/3, (7 pi)/4#

Given: #tan^2 x + tan x = sqrt(3) tan x + sqrt(3)," in "[0, 2 pi]#

Rearrange the equation to be #= 0#:

#tan^2 x + tan x - sqrt(3) tan x - sqrt(3) = 0#

Group factor:

#(tan^2 x + tan x) + (- sqrt(3) tan x - sqrt(3)) = 0#

#tan x ( tan x + 1) -sqrt(3) (tan x + 1) = 0#

#(tan x + 1)(tan x - sqrt(3)) = 0#

#tan x = -1; " "tan x = sqrt(3)#

The tangent is positive in quadrants I, III and negative in quadrants II & IV.

#x = (3 pi)/4, (7 pi)/4; " " x = pi/3, (4 pi)/3#