Question

Question #946b8

Answer 1

`x=cos^-1[3/4+-sqrt(17)/4]+2pin` with `ninZZ`

Explanation:

Transform the given equation into polynomial form using the fact that `cos(2x)=2cos^2(x)-1`

`cos(2x)-3cos(x)=0`
`2cos^2(x)-1-3cos(x)=0`
`2cos^2(x)-3cos(x)-1=0`

Write this quadratic equation in standard form by dividing by 2
`cos^2(x)-3/2cos(x)-1/2=0`

Solve the quadratic equation by completing the square. Add 1/2 to both sides
`cos^2(x)-3/2cos(x)=1/2`

Take half the coefficient of `cos(x)` and square it, then add to both sides.
`cos^2(x)-3/2cos(x)+9/16=17/16`

Factor the left-hand side.
`(cos(x)-3/4)^2=17/16`

`cos(x)-3/4=+-sqrt(17)/4`

`cos(x)=3/4+-sqrt(17)/4`

`cos^-1[cos(x)]=cos^-1[3/4+-sqrt(17)/4]+2pin` with `ninZZ`
`x=cos^-1[3/4+-sqrt(17)/4]+2pin` with `ninZZ`

Answer 2

An alternative and a few details.

Explanation:

One might also use the Quadratic Formula.
After the step

`2cos^2(x) - 3cosx - 1 = 0`, note that the equation is quadratic in form -- much like `2y^2-3y-1=0`.

Therefore
`cosx = (3 +-sqrt(9 - 4(2)(-1)))/2(2)`
`cosx = (3 +-sqrt(9 + 8))/4`
`cosx = (3 +-sqrt(17))/4`

Now observe that `3 + sqrt(17) > 3 + 4 > 4`.
Therefore `(3 + sqrt(17))/4 > 1`, which means that cosx never attains that value.

However `(3 - sqrt(17))/4` lies between -1 and 0. Therefore

`cosx = (3 - sqrt(17))/4` for some values of x in the second and third quadrants.

When we write `Cos^(-1)u` of any value u, we mean the angle in the first or second quadrant whose cosine is "u."

Thus `x = Cos^(-1)((3 - sqrt(17))/4)` is in Quadrant II, and corresponding to it, for every integer n, `x = Cos^(-1)((3 - sqrt(17))/4) + 2npi` is a solution to the equation.

In the third quadrant, the angle with the same reference angle as `x_1 = Cos^(-1)((3 - sqrt(17))/4)` is `x_2 = 2pi - Cos^(-1)((3 - sqrt(17))/4)`

The solutions to the equation are
`x = Cos^(-1)((3 - sqrt(17))/4) + 2npi`
and
`x = -Cos^(-1)((3 - sqrt(17))/4) + 2npi`

Since cosine is an even function, [`cos(theta) = cos(-theta)` for all `theta`], this is the sort of answer that we ought to expect.