# What is x if 2*sin x + 1 = tan x + 2*sin x * tan x, if x lies in [0; 2*pi] ?

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We have to find x so that 2*sin x + 1 = tan x + 2*sin x * tan x, where x is in the interval [ 0 ; 2pi]

2*sin x + 1 = tan x + 2*sin x * tan x

=> 2*sin x - 2*sin x * tan x = tan x - 1

=> 2*sin x( 1 - tan x) = -( 1- tan x)

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

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

=> tan x = 1 and 2*sin x = -1

=> tan x = 1 and sin x = -1/2

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

=> x = 45 degree and 330 degree.

**Therefore x= 45 degree and 330 degree.**

We'll use the fact that the function tangent is a ratio:

tan x = sin x/cos x

We'll re-write the given equation moving all terms to one side:

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

We'll multiply by cos x:

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

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

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

We'll factorize by 2 sin x + 1:

(2 sin x + 1)(cos x - sin 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:

cos x - sin x = 0

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

1 - tan x = 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 interval[0 , 2pi], are: {pi/4 ; 5pi/4 ; 7pi/6 ; 11pi/6}.**