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...

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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.**