Question

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

Solve the equation for 0 ≤ x ≤ 2π.

Answer 1

`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`

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`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

`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`