For an angle to be considered a special angle, the angle must be able to produce a $\displaystyle 30-60-90$ or a $\displaystyle 45-45-90$ triangle.  

The only angle that is not capable of the special angles formation is $\displaystyle 75 \textup{ degrees}$.

We first make the substitution $\displaystyle k=5x$.

In the interval $\displaystyle 0\leq k\leq \pi$, the equation $\displaystyle \cos k=\frac{-1}{2}$ has the solution $\displaystyle k=\frac{2\pi}{3}$.

Solving for $\displaystyle x$,

$\displaystyle \begin{align*} 5x&=\frac{2\pi}{3}\\ x&=\frac{2\pi}{15} \end{align*}$.

From the unit circle, the value of 

$\displaystyle \sin(\pi/6)=1/2$.

This can be found using the coordinate pair associated with the angle $\displaystyle \pi/6$ which is $\displaystyle (\sqrt3/2,1/2)$.

Recall that the $\displaystyle (x,y)$ pair are $\displaystyle (\cos,\sin)$.

Recall that the unit circle can be broken down into four quadrants. Each quadrant has similar coordinate pairs basic on the angle. The only difference between the actual coordinate pairs is the sign on them. In quadrant I all signs are positive. In quadrant II only sine and cosecant are positive. In quadrant III tangent and cotangent are positive and quadrant IV only cosine and secant are positive.

$\displaystyle \frac{11\pi}{6}$ has the reference angle of $\displaystyle \frac{\pi}{6}$ and lies in quadrant IV therefore $\displaystyle (cos, sin)=(+,-)$.

From the unit circle, the coordinate point of 

$\displaystyle (\sqrt3/2,-1/2)$ corresponds with the angle $\displaystyle \frac{11\pi}{6}$.  

Locate $\displaystyle \frac{5\pi}{6}$ on the unit circle.

Sine is related to the why y value of the coordinate point because it is opposite/Hyp.

In other words the $\displaystyle (x,y)$ pair of the point located on the unit circle that extends from the origin is $\displaystyle (\cos(\theta),\sin(\theta))$.

The coordinate pair for $\displaystyle \frac{5\pi}{6}$ is $\displaystyle \left(\frac{-\sqrt{3}}{2},\frac{1}{2}\right)$ thus,

 $\displaystyle \sin\frac{5\pi}{6}=\frac{1}{2}$

Locate $\displaystyle \frac{4\pi}{3}$ on the unit circle.

Cosine is related to the why x value of the points, because it is Adj/Hyp.

In other words the $\displaystyle (x,y)$ pair of the point located on the unit circle that extends from the origin is $\displaystyle (\cos(\theta),\sin(\theta))$.

The coordinate point of $\displaystyle \frac{4\pi}{3}$ is $\displaystyle \left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right )$ thus,

$\displaystyle \cos \frac{4\pi}{3}=-\frac{1}{2}$.

Use the unit circle to locate the following:

$\displaystyle \frac{2\pi}{3}$ 

Recall that the $\displaystyle (x,y)$ pair of the point located on the unit circle that extends from the origin is the following:

$\displaystyle (\cos(\theta),\sin(\theta))$ 

Since the tangent is equal to the sine divided by the cosine, we need to find the following:

$\displaystyle \sin \frac{2\pi}{3}=\frac{\sqrt{3}}{2}$

$\displaystyle \cos \frac{2\pi}{3}=-\frac{1}{2}$

Now, we will use this information to solve the problem. 

$\displaystyle \tan \frac{2\pi}{3}=\frac{\sin\frac{2\pi}{3}}{\cos \frac{2\pi}{3}}$

Substitute the calculated values:

$\displaystyle \frac{\sin\frac{2\pi}{3}}{\cos \frac{2\pi}{3}}=\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}$

Solve. Remember that dividing by a fraction is the same as multiplying by its reciprocal.

$\displaystyle \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=\frac{\sqrt{3}}{2}\times-\frac{2}{1}$

$\displaystyle \frac{\sqrt{3}}{\not{2}}\times-\frac{\not{2}}{1}=-\sqrt{3}$