# √(cosx)=2cosx-1solve the equation

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Solve for x: `sqrt(cosx)=2cosx-1` Square both sides:

`cosx=4cos^2x-4cosx+1`

`4cos^2x-5cosx+1=0`

`(4cosx-1)(cosx-1)=0`

`cosx=1/4` or cosx=1

However, since we squared both sides we must check for extraneous solutions:

If `cosx=1/4` then `sqrt(cosx)=sqrt(1/4)=1/2` and 2cosx-1=`2(1/4)-1=-1/2` so this is an extraneous solution.

If cosx=1 then `sqrt(cosx)=1` and 2cosx-1=2(1)-1=1 which works.

cosx=1==>` ` `x=0+-2kpi`

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The solutions to `sqrt(cosx)=2cosx-1` are `x=0+-2kpi,k in ZZ`

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The graph of `sqrt(cosx)` in red; the graph of 2cosx-1 in black:

** The straight line portions of the red graph are artifacts of the grapher: you cannot take the square root of a negative number in the reals, so anywhere the cosine is negative is not in the domain.

You need to raise to square both sides, such that:

`(sqrt(cos x)) = (2cos x - 1)^2`

Expanding the square yields:

`cos x = 4cos^2 x - 4 cos x + 1`

Moving all the terms to one side yields:

`4cos^2 x - 4 cos x - cos x + 1 = 0`

`4cos^2 x - 5 cos x + 1 = 0`

You need to substitute t for `cos x` such that:

`4t^2 - 5t + 1 = 0`

You need to split the term `-5t` in two terms such that :

`4t^2 - 4t - t + 1 = 0`

Grouping the terms yields:

`(4t^2 - 4t) - (t - 1) = 0`

Factoring out `4t` in the first group yields:

`4t(t - 1) - (t - 1) = 0`

Factoring out `(t-1)` yields:

`(t-1)(4t-1) = 0 => {(t-1=0),(4t-1=0):} => {(t=1),(t=1/4):}`

Substituting back `cos x` for t yields:

`cos x = 1 > 0 => x = 2npi`

`cos x = 1/4 > 0 => x = +-cos^(-1)(1/4) + 2npi`

**Hence, evaluating the solutions to the given equation yields `x = 2npi` and **`x = +-cos^(-1)(1/4) + 2npi.`