# Question dcf23

Jun 8, 2016

theta=60˚,109.4712˚

#### Explanation:

We have the equation

$6 {\cos}^{2} \theta - \cos \theta = 1$

Notice that this is similar to a quadratic equation--we have a variable expression being squared, a variable expression to the first power, and a constant. We should try to solve this as we normally would a quadratic first and then bring the trigonometry into play.

To make this more approachable, a good (although not necessary method) is to let $u = \cos \theta$. Then, we have the equation

$6 {u}^{2} - u - 1 = 0$

To factor this quadratic, find a pair of factors of $- 6$ whose sum is $- 1$: the numbers $- 3$ and $2$.

$6 {u}^{2} - 3 u + 2 u - 1 = 0$

$3 u \left(2 u - 1\right) + 1 \left(2 u - 1\right) = 0$

$\left(3 u + 1\right) \left(2 u - 1\right) = 0$

$\left\{\begin{matrix}3 u + 1 = 0 \\ 2 u - 1 = 0\end{matrix}\right.$

$\left\{\begin{matrix}u = - \frac{1}{3} \\ u = \frac{1}{2}\end{matrix}\right.$

Now, since $u = \cos \theta$,

$\left\{\begin{matrix}\cos \theta = - \frac{1}{3} \\ \cos \theta = \frac{1}{2}\end{matrix}\right.$

Solving just for the first equation:

$\cos \theta = - \frac{1}{3}$

Using the inverse cosine function,

$\theta = \arccos \left(- \frac{1}{3}\right)$

Plugging this into a calculator yields theta=109.4712˚ (approximately).

The second equation is a commonly known value:

$\cos \theta = \frac{1}{2}$

theta=60˚#