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

Solve the equation for 0 ≤ x ≤ 2π.enter image source here

2 Answers
Jul 25, 2018

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

Explanation:

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

Jul 25, 2018

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

Explanation:

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