# Solve inequality (cos x-sin x)/(2cos x+1 )> 0r equal 0

*print*Print*list*Cite

### 4 Answers

`ERRATA-CORRIGE:`

`x<= -3/4 pi` `-2/3 pi < x <2/3 pi` `x>=3/4 pi`

The points of value `+- 2/3 pi ` are to be discarged for function is not defined there.

`(cosx-sinx)/(2cosx+1) >=0`

`cosx -sinx >=0` `tan x<=1` `-3/4 pi<= x<= pi/4`

On the other side:

`2cosx+1>=0` `cosx>= -1/2` `-2/3pi <= x<=2/3 pi`

Now,since inequality is request for argument `>=0` we have to find intervals where both the part of ratioare or negative or positive.

Let you see:

`cosx-sinx` : `----- (-3/4 pi) ++++ (-2/3 pi) +++++ (2/3 pi) ++++(3/4 pi) ------`

`2cosx +1` :

`-----(-3/4 pi)-----(-2/3pi)+++++(2/3pi)-----(3/4pi)-----`

since `+ xx+=+` and `- xx - =+`

Solutions are

`x<= -3/4 pi` ; `-2/3pi <=x <=2/3 pi` ;`x>= 3/4 pi`

We have

`{cos(x)-sin(x)}/(2cos(x)+1)>0`

If `2cos(x)+1>0and cos(x)-sin(x)>0`

`cos(x) > -1/2`

`cos(x)>cos(180-60)`

`cos(x)>cos(120^o)`

`x < (2pi)/3 --------------------(i)`

`` `cos(x)>sin(x)`

`sin(pi/2-x) > sin(x)`

`pi/2-x > x`

`pi/2 >2x`

`x < pi/4` (ii)

Thus from (i) and (ii)

we have `x < pi/4`

Now if

cos(x)-sin(x) < 0 and 2cos(x)+1 <0

cos(x)< sin(x)

sin(pi/2-x)< sin(x)

(pi/2)-x < x

pi/2 < 2x

pi/4 < x (iii)

2cos(x)+1 <0

2cos(x)< -1

cos(x) <-1/2

cos(x)< cos(pi-pi/3)

x>(2 pi )/3 (iv)

from (iii) and (iv), we do not find any value which satisfy both inequalitie.

Thus solution of the problem is

`0<=x<pi/4`