# What value(s) of x between 0 and 2`pi` solve the following equation: `cos^2(x) - cos x - 6 = 0`

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You need to come up withthe following substitution, such that:

`cos x = y => cos^2 x = y^2`

Replacing the variable yields:

`y^2 - y - 6 = 0`

You may use quadratic formula to solve for y the equation, such that:

`y_(1,2) = (-b+-sqrt(b^2 - 4ac))/(2a)`

Identifying the coefficients a,b,c yields:

`a=1, b=-1, c=-6`

`y_(1,2) = (1+-sqry(1 + 24))/2 => y_(1,2) = (1+-5)/2`

`y_1 = 3 ; y_2 = -2`

You need to solve for x the following equations, such that:

`cos x = y_1 => cos x = 3` invalid since `cos x in [-1,1]`

`cos x = y_2 => cos x = -2` invalid since `cos x in [-1,1]`

**Hence, evaluating the solutions to the given equation yields that there are no solutions **`x in [0,2pi].`

`cos^2x - cosx -6 = 0`

`cos^2x -2 1/2 cosx +1/4 -25/4=0 `

`(cosx -1/2) ^2 =25/4`

`cos x - 1/2 = +- 5/2`

`cosx= 1/2 +- 5/2` `cosx = 3` `cosx= -2`

both value not accepted