# 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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We have to solve for x in the equation 2 sin x tan x + tan x - 2 sin x - 1= 0 for x lying between 0 and 2*pi.

2 sinx tan x + tan x - 2 sin x - 1= 0

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

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

For tan x - 1 =0

tan x = 1

x = arc tan 1 = pi/ 4.

For 2 sin x + 1 = 0

=> sin x = -1/2

=> x = arc sin (-1/2)

=> x = 7pi / 6, or 11 pi / 6

**Therefore x is equal to pi/4, 7pi/6 or 11 pi/6.**

We'll write the function tan x = sin x/cos x

We'll re-write the given equation:

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 set the first factor as zero:

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

To solve for x in the equation 2sin xtan x + tan x -2sinx-1=0

The LHS of the the equation could be grouped and factorised as below:

tanx(2sinx+1)-1(2sinx+1)

(2sinx+1)(tanx-1).

Therefore the given equation could be written asbelow:

(2sinx+1)(tanx-1) = 0. So zero is a product of 2 factors.So set either of thfactors to zero.

2six+1 = 0 gives: six = -1/2, or x = 5pi/8, or 7pi/6, or 11pi/6

tanx-1 = 0 givex tanx = 1, So x = pi/4, or 3pi/4.