# Given that 2sinx+1=2sin^2x/cosx+sinx/cosx what is x.Given that 2sinx+1=2sin^2x/cosx+sinx/cosx what is x.

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You need to factor out `sin x/cos x` to the right side, such that:

`2 sin x + 1 = (sin x/cos x)(2 sin x + 1)`

Subtracting the product `(sin x/cos x)(2 sin x + 1)` both sides yields:

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

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

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

Replacing the tanget function `tan x` for the quotient `sin x/cos x` yields:

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

Using zero product rule yields:

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

`2 sin x + 1 = 0 => 2 sin x = -1 => sin x = -1/2 => x = (-1)^(n+1)(pi/6) + npi`

`1 - tan x = 0 => tan x = 1 => x = pi/4 + npi`

**Hence, evaluating the solutions to the given equation yields **`x = = (-1)^(n+1)(pi/6) + npi, x = pi/4 + npi.`

We'll re-write the given equation and we'll multiply by cos x, having all terms to one side:

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 put 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 put the next 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, in the interval[0 , 2pi], are: {pi/4 ; 5pi/4 ; 7pi/6 ; 11pi/6}.**