# Solve: cos^2x - 3/4 = 0Solve exactly for special angles, to 2 decimal places otherwise.

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Solve `cos^2x-3/4=0` :

`cos^2x=3/4`

`cosx=+-sqrt(3/4)`

`cosx=+-(sqrt(3))/2`

`x=cos^(-1)(+-sqrt(3)/2)`

(1)`cos^(-1)sqrt(3)/2=pi/6` . Note that the domain for the arccosine is `0<x<pi` ; the other angle in `0<x<2pi` where `cosx=sqrt(3)/2` is `(11pi)/6`

(2) `cos^(-1)(-sqrt(3)/2)=(5pi)/6` . Again there is another angle in `0<x<2pi` where `cosx=-sqrt(3)/2` and that is `x=(7pi)/6`

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**The solution to `cosx-3/4=0` is `x=pi/6+npi,x=(5pi)/6+npi` for `n in ZZ` (n an integer).**

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You need to keep the squared cosine to the left such that:

`cos^2 x = 3/4`

You should take square root both sides such that:

`cos x = +-sqrt(3/4)`

Taking the positive value, `cos x = sqrt3/2` , yields:

`cos x = sqrt3/2 =gt x = +-cos^(-1)(sqrt3/2) + 2npi`

`x = +-pi/6 + 2npi`

Taking the negative value, `cos x = -sqrt3/2` , yields:

`x = -+pi/6 + 2npi`

You should remember that cosine function is positive in the quadrants 1 and 4 and it has negative values in quadrants 2 and 3.

**Hence, evaluating the general solutions to the given equation yields `x = +-pi/6 + 2npi ` or `x = -+pi/6 + 2npi.` **