# Question #946b8

May 14, 2017

$x = {\cos}^{-} 1 \left[\frac{3}{4} \pm \frac{\sqrt{17}}{4}\right] + 2 \pi n$ with $n \in \mathbb{Z}$

#### Explanation:

Transform the given equation into polynomial form using the fact that $\cos \left(2 x\right) = 2 {\cos}^{2} \left(x\right) - 1$

$\cos \left(2 x\right) - 3 \cos \left(x\right) = 0$
$2 {\cos}^{2} \left(x\right) - 1 - 3 \cos \left(x\right) = 0$
$2 {\cos}^{2} \left(x\right) - 3 \cos \left(x\right) - 1 = 0$

Write this quadratic equation in standard form by dividing by 2
${\cos}^{2} \left(x\right) - \frac{3}{2} \cos \left(x\right) - \frac{1}{2} = 0$

Solve the quadratic equation by completing the square. Add 1/2 to both sides
${\cos}^{2} \left(x\right) - \frac{3}{2} \cos \left(x\right) = \frac{1}{2}$

Take half the coefficient of $\cos \left(x\right)$ and square it, then add to both sides.
${\cos}^{2} \left(x\right) - \frac{3}{2} \cos \left(x\right) + \frac{9}{16} = \frac{17}{16}$

Factor the left-hand side.
${\left(\cos \left(x\right) - \frac{3}{4}\right)}^{2} = \frac{17}{16}$

$\cos \left(x\right) - \frac{3}{4} = \pm \frac{\sqrt{17}}{4}$

$\cos \left(x\right) = \frac{3}{4} \pm \frac{\sqrt{17}}{4}$

${\cos}^{-} 1 \left[\cos \left(x\right)\right] = {\cos}^{-} 1 \left[\frac{3}{4} \pm \frac{\sqrt{17}}{4}\right] + 2 \pi n$ with $n \in \mathbb{Z}$
$x = {\cos}^{-} 1 \left[\frac{3}{4} \pm \frac{\sqrt{17}}{4}\right] + 2 \pi n$ with $n \in \mathbb{Z}$

May 15, 2017

An alternative and a few details.

#### Explanation:

One might also use the Quadratic Formula.
After the step

$2 {\cos}^{2} \left(x\right) - 3 \cos x - 1 = 0$, note that the equation is quadratic in form -- much like $2 {y}^{2} - 3 y - 1 = 0$.

Therefore
$\cos x = \frac{3 \pm \sqrt{9 - 4 \left(2\right) \left(- 1\right)}}{2} \left(2\right)$
$\cos x = \frac{3 \pm \sqrt{9 + 8}}{4}$
$\cos x = \frac{3 \pm \sqrt{17}}{4}$

Now observe that $3 + \sqrt{17} > 3 + 4 > 4$.
Therefore $\frac{3 + \sqrt{17}}{4} > 1$, which means that cosx never attains that value.

However $\frac{3 - \sqrt{17}}{4}$ lies between -1 and 0. Therefore

$\cos x = \frac{3 - \sqrt{17}}{4}$ for some values of x in the second and third quadrants.

When we write $C o {s}^{- 1} u$ of any value u, we mean the angle in the first or second quadrant whose cosine is "u."

Thus $x = C o {s}^{- 1} \left(\frac{3 - \sqrt{17}}{4}\right)$ is in Quadrant II, and corresponding to it, for every integer n, $x = C o {s}^{- 1} \left(\frac{3 - \sqrt{17}}{4}\right) + 2 n \pi$ is a solution to the equation.

In the third quadrant, the angle with the same reference angle as ${x}_{1} = C o {s}^{- 1} \left(\frac{3 - \sqrt{17}}{4}\right)$ is ${x}_{2} = 2 \pi - C o {s}^{- 1} \left(\frac{3 - \sqrt{17}}{4}\right)$

The solutions to the equation are
$x = C o {s}^{- 1} \left(\frac{3 - \sqrt{17}}{4}\right) + 2 n \pi$
and
$x = - C o {s}^{- 1} \left(\frac{3 - \sqrt{17}}{4}\right) + 2 n \pi$

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