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

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

Solve the equation for 0 ≤ x ≤ 2π.

2 Answers

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Answer 1 · 2018-07-25T06:08:55

$x = \frac{π}{3}, \frac{4 π}{3}, \frac{3 π}{4}, \frac{7 π}{4}$ $x=pi/3, (4pi)/3,(3pi)/4, (7pi)/4$

Explanation:

${tan}^{2} x + tan x = \sqrt{3} tan x + \sqrt{3}$ $tan^2x+tanx=sqrt3 tanx+sqrt3$

${tan}^{2} x + tan x - \sqrt{3} tan x - \sqrt{3} = 0$ $tan^2x+tanx-sqrt3tanx-sqrt3=0$

$tan x (tan x + 1) - \sqrt{3} (tan x + 1) = 0$ $tanx(tanx+1)-sqrt3(tanx+1)=0$

$(tan x - \sqrt{3}) (tan x + 1) = 0$ $(tanx-sqrt3)(tanx+1)=0$

$tan x - \sqrt{3} = 0$ $tanx-sqrt3=0$ or $tan x + 1 = 0$ $tanx+1=0$

$tan x - \sqrt{3} = 0$ $tanx-sqrt3=0$
$tan x = \sqrt{3}$ $tanx=sqrt3$
$x = \frac{π}{3}, π + \frac{π}{3}$ $x=pi/3, pi+pi/3$ --> $tan x$ $tan x$ is positive in the first and third quadrant
$x = \frac{π}{3}, \frac{4 π}{3}$ $x=pi/3, (4pi)/3$

$tan x + 1 = 0$ $tanx+1=0$
$tan x = - 1$ $tanx=-1$
$x = π - \frac{π}{4}, 2 π - \frac{π}{4}$ $x=pi-pi/4, 2pi-pi/4$ --> $tan x$ $tanx$ is negative in the second and fourth quadrant
$x = \frac{3 π}{4}, \frac{7 π}{4}$ $x=(3pi)/4, (7pi)/4$

Answer 2 · 2018-07-25T06:12:17

$\frac{π}{3}, \frac{3 π}{4}, \frac{4 π}{3}, \frac{7 π}{4}$ $pi/3, (3 pi)/4, (4 pi)/3, (7 pi)/4$

Explanation:

Given: ${tan}^{2} x + tan x = \sqrt{3} tan x + \sqrt{3}, in [0, 2 π]$ $tan^2 x + tan x = sqrt(3) tan x + sqrt(3)," in "[0, 2 pi]$

Rearrange the equation to be $= 0$ $= 0$ :

${tan}^{2} x + tan x - \sqrt{3} tan x - \sqrt{3} = 0$ $tan^2 x + tan x - sqrt(3) tan x - sqrt(3) = 0$

Group factor:

$({tan}^{2} x + tan x) + (- \sqrt{3} tan x - \sqrt{3}) = 0$ $(tan^2 x + tan x) + (- sqrt(3) tan x - sqrt(3)) = 0$

$tan x (tan x + 1) - \sqrt{3} (tan x + 1) = 0$ $tan x ( tan x + 1) -sqrt(3) (tan x + 1) = 0$

$(tan x + 1) (tan x - \sqrt{3}) = 0$ $(tan x + 1)(tan x - sqrt(3)) = 0$

$tan x = - 1; tan x = \sqrt{3}$ $tan x = -1; " "tan x = sqrt(3)$

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

$x = \frac{3 π}{4}, \frac{7 π}{4}; x = \frac{π}{3}, \frac{4 π}{3}$ $x = (3 pi)/4, (7 pi)/4; " " x = pi/3, (4 pi)/3$

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

Solve the equation for 0 ≤ x ≤ 2π.

2 Answers

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