Solve for x in the equation: 2 sin x tan x + tan x - 2 sin x - 1 = 0 for 0<=x<=2pi.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You also may use factorization as alternative method, such that:

`(2 sin x tan x + tan x) - (2 sin x + 1) = 0`

`tan x(2 sin x + 1) - (2 sin x + 1) = 0`

Factoring out `(2 sin x + 1)` yields:

`(2 sin x + 1)(tan x - 1) = 0 => {(2 sin x + 1 = 0),(tan x - 1 = 0):}`

`{(2 sin x = -1),(tan x= 1):} => {(sin x = -1/2),(tan x= 1):}`

`{(x = sin^(-1)(-1/2)),(x = tan^(-1) 1):}`

`{(x = pi + pi/6),(x = pi + pi/4):} => {(x = (7pi)/6),(x = (5pi)/4):}`

`{(x = 2pi - pi/6),(x = pi/4):} => {(x = (11pi)/6),(x = pi/4):}`

Hence, evaluating the solutions to the given equation yields `x = (7pi)/6, x = (5pi)/4, x = (11pi)/6, x = pi/4.`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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tan x = sin x/cos x

2 sin x (sin x/cos x) + sin x/cos x - 2 sin x - 1 = 0

We'll multiply by cos x:

2(sin x)^2 + sin x - 2sin x*cos x - cos x = 0

We'll factorize the first 2 terms by sin x and the last 2 terms by - cos x:

sin x(2 sin x + 1) - cos x(2 sin x + 1) = 0

We'll factorize by 2 sin x + 1:

(2 sin x + 1)(sin x - cos x) = 0

We'll cancel the first factor:

2 sin x + 1 = 0

We'll subtract 1;

2sinx = -1

sin x = -1/2

x = arcsin (-1/2)

The sine function is negative in the 3rd and 4th quadrants:

x = pi + pi/6

x = 7pi/6 (3rd qudrant)

x = 2pi - pi/6

x = 11pi/6 (4th qudrant)

We'll set the other factor as zero:

sin x - cos x = 0

This is an homogeneous equation and we'll divide it by cos x:

tan x - 1 = 0

tan x = 1

The function tangent is positive in the 1st and the 3rd qudrants:

x = arctan 1

x = pi/4 (1st quadrant)

x = pi+ pi/4

x = 5pi/4 (3rd qudrant)

The set of solutions of the equation, over the interval [0 , 2pi], are: {pi/4 ; 5pi/4 ; 7pi/6 ; 11pi/6}.

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