solve for x, where 0<x<2pi. 2cos^2 (x)+cos (x)=1

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`2cos^2 x-1+cos x=0`

`2cos^2 x -1` reminds me of a classical formula `2cos^2 x -1=cos 2x`

Our equation is equivalent to `cos 2x+cos x=0`

`cos 2x=-cos x=cos (pi-x)`

iff `2x=pi-x +2k pi ` or 2x=-(pi-x)+2kpi for some  `kin ZZ `

iff `3x=pi+2k pi ` or `x=-pi+2kpi` for some `k in ZZ`

iff `x=pi/3+2k pi/3` or `x=-pi +2k pi ` for some `k in ZZ`

iff `x=pi/3 (k=0)` ,

`x=pi/3+2pi/3=pi (k=1)` ,

`x=pi/3+4pi/3=5pi/3 ` `(k=2), ` 

or `x=pi (k=1)` .


Any other values of k will provide solutions outside

`[0,2pi]. `


Solutions `x=pi/3, x=pi, x=5pi/3`


Second method:

Let's X be cos x

The equation is `2X^2+X=1`

iff `2x^2+X-1=0 `

iff `(X+1)(2X-1)=0`

iff `X=-1` or `X=1/2`

iff `cos x=-1` or `cos =-1/2`

In `[0,2pi] `

iff `x=pi` or `x=pi/3` or `x=5pi/3`




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