# Question b6d43

May 30, 2017

$\frac{5 \pi}{6}$ and $\frac{7 \pi}{6}$

#### Explanation:

$\cos \left(x\right) = - \frac{\sqrt{3}}{2}$

If we apply the inverse of $\cos$, we can isolate $x$ and use a calculator to solve:

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

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

$x = \frac{5 \pi}{6}$

My calculator only gave me one answer, but that may not be correct....

Now, let's try not using a calculator, and use our logic

In trigonometry, there are some ratios that keep popping up:
$\frac{\sqrt{3}}{2}$, $\frac{1}{2}$, and $\frac{\sqrt{2}}{2}$

Let's focus on $\frac{\sqrt{3}}{2}$:

$\sin \left({60}^{o}\right)$ will give you $\frac{\sqrt{3}}{2}$, or $\sin \left(\frac{\pi}{3}\right)$.

Sine and Cosine are supplemantary, so if $\sin \left(60\right)$ gives you $\frac{\sqrt{3}}{2}$, $\cos \left(90 - 60\right)$ will give you the same answer.

$C o s \left({30}^{o}\right)$ or $\cos \left(\frac{\pi}{6}\right)$ equals $\frac{\sqrt{3}}{2}$. Now we have a referance angle for our problem.

Now, we need to find the angle that gill give us the ratio $- \frac{\sqrt{3}}{2}$. To find that, we need to find the quadrant that gives us a negative number:

$\textcolor{w h i t e}{- - - -} \textcolor{w h i t e}{S} \textcolor{w h i t e}{- -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- -} \textcolor{w h i t e}{A} \textcolor{w h i t e}{- - - -}$
$\textcolor{w h i t e}{- - - -} \textcolor{w h i t e}{S} \textcolor{w h i t e}{- -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- -} \textcolor{w h i t e}{A} \textcolor{w h i t e}{- - - -}$
$\textcolor{w h i t e}{- - - -} \textcolor{red}{S} \textcolor{w h i t e}{- -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- -} \textcolor{red}{A} \textcolor{w h i t e}{- - - -}$
$\textcolor{w h i t e}{- - - -} \textcolor{w h i t e}{S} \textcolor{w h i t e}{- -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- -} \textcolor{w h i t e}{A} \textcolor{w h i t e}{- - - -}$
$\textcolor{b l a c k}{- - - -} \textcolor{w h i t e}{S} \textcolor{b l a c k}{- -} \textcolor{b l a c k}{|} \textcolor{b l a c k}{- -} \textcolor{w h i t e}{A} \textcolor{b l a c k}{- - - -}$
$\textcolor{w h i t e}{- - - -} \textcolor{w h i t e}{S} \textcolor{w h i t e}{- -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- -} \textcolor{w h i t e}{A} \textcolor{w h i t e}{- - - -}$
$\textcolor{w h i t e}{- - - -} \textcolor{w h i t e}{S} \textcolor{w h i t e}{- -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- -} \textcolor{w h i t e}{A} \textcolor{w h i t e}{- - - -}$
$\textcolor{w h i t e}{- - - -} \textcolor{red}{T} \textcolor{w h i t e}{- -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- -} \textcolor{red}{C} \textcolor{w h i t e}{- - - -}$
$\textcolor{w h i t e}{- - - -} \textcolor{w h i t e}{S} \textcolor{w h i t e}{- -} \textcolor{b l a c k}{|} \textcolor{w h i t e}{- -} \textcolor{w h i t e}{A} \textcolor{w h i t e}{- - - -}$

$\textcolor{red}{A}$ tells us that All functions are positive in this quadrant

$\textcolor{red}{S}$ tells us that Sine functions are positive in this quadrant

$\textcolor{red}{T}$ tells us that Tangent functions are positive in this quadrant

$\textcolor{red}{C}$ tells us that Cosine functions are positive in this quadrant

I remember this using the mnemonic All Students Take Calculus

We're looking for $\cos \left(\theta\right) = \textcolor{red}{-} \frac{\sqrt{3}}{2}$. That means we need to know the quadrants where the rations are negative for $\cos$.

The quadrants we are looking for are Q II and III

If we go around the circle by $\frac{\pi}{6}$:

QI

$\textcolor{b l u e}{\frac{\pi}{6}}$
$\frac{2 \pi}{6}$ or $\textcolor{b l u e}{\frac{\pi}{3}}$

QII

$\frac{3 \pi}{6}$ or $\textcolor{b l u e}{\frac{\pi}{2}}$
$\frac{4 \pi}{6}$ or color(blue)((2pi)/30#
$\textcolor{b l u e}{\frac{5 \pi}{6}}$

QIII

$\frac{6 \pi}{6}$ or $\textcolor{b l u e}{\pi}$
$\textcolor{b l u e}{\frac{7 \pi}{6}}$
$\frac{8 \pi}{6}$ or $\textcolor{b l u e}{\frac{4 \pi}{3}}$

QIV

$\frac{9 \pi}{6}$ or $\textcolor{b l u e}{\frac{3 \pi}{2}}$
$\frac{10 \pi}{6}$ or $\textcolor{b l u e}{\frac{5 \pi}{3}}$
$\textcolor{b l u e}{\frac{11 \pi}{6}}$
$\frac{12 \pi}{6}$ or $\textcolor{b l u e}{2 \pi}$

The two solutions we care about are in Q II and III and are based on the reference angle $\frac{\pi}{6}$

The only ones that follow those restraints are $\frac{5 \pi}{6}$ and $\frac{7 \pi}{6}$

And just to check that both values are valid, let's use a calulator to make sure that $\cos \left(\frac{5 \pi}{6}\right)$ and $\cos \left(\frac{7 \pi}{6}\right)$ both give us $- \frac{\sqrt{3}}{2}$

It does! So, both solutions are correct. Good work!

May 31, 2017

$\frac{5 \pi}{6}$
$\frac{7 \pi}{6}$

#### Explanation:

Solve: $\cos x = - \frac{\sqrt{3}}{2}$
Trig table and unit circle give 2 solutions:
$x = \pm \frac{5 \pi}{6} + 2 k \pi$
The arc $- \frac{5 \pi}{6}$ is co-terminal to the arc $\left(2 \pi - \frac{5 \pi}{6}\right) = \frac{7 \pi}{6}$
$x = \frac{5 \pi}{6} + 2 k \pi$
$x = \frac{7 \pi}{6} + 2 k \pi$