First lets identify the angles that make up the third quadrant. Quadrant three is $\displaystyle 180^\circ$ to $\displaystyle 270^\circ$ or in radians, $\displaystyle \pi$ to $\displaystyle \frac{3\pi}{2}$ thus, any angle that does not fall within this range is not in quadrant three.

Therefore, the correct answer,

$\displaystyle \small \frac{7\pi}{3}$ is not in quadrant three because it is in the first quadrant.

This is clear when we subtract

$\displaystyle \small \frac{7\pi}{3} - 2 \pi = \frac{7 \pi}{3} - \frac{6\pi}{3} = \frac{\pi }{3}$.

First lets identify the different quadrants.

Quadrant I:$\displaystyle 0\rightarrow \frac{\pi}{2}$

Quadrant II: $\displaystyle \frac{\pi}{2}\rightarrow \pi$

Quadrant III: $\displaystyle \pi \rightarrow \frac{3\pi}{2}$

Quadrant IV: $\displaystyle \frac{3\pi}{2}\rightarrow 2\pi$

Now looking at our possible answer choices, we will add or subtract $\displaystyle 2\pi$ until we get the reduced fraction of the angle. This will tell us which quadrant the angle lies in.

$\displaystyle \small \frac{-2\pi}{3}+2\pi=\frac{-2\pi}{3}+\frac{6\pi}{3}=\frac{4\pi}{3}$ thus in quadrant III. 

$\displaystyle \small \frac{13\pi}{4}-2\pi=\frac{13\pi}{4}-\frac{8\pi}{4}=\frac{5\pi}{4}$ thus in quadrant III.

Therefore,

$\displaystyle \small \frac{-2\pi}{3}$ and $\displaystyle \small \frac{13\pi}{4}$ is the correct answer.

First lets identify the different quadrants.

Quadrant I:$\displaystyle 0\rightarrow \frac{\pi}{2}$

Quadrant II: $\displaystyle \frac{\pi}{2}\rightarrow \pi$

Quadrant III: $\displaystyle \pi \rightarrow \frac{3\pi}{2}$

Quadrant IV: $\displaystyle \frac{3\pi}{2}\rightarrow 2\pi$

The correct answer,$\displaystyle \small \frac{-4\pi}{3}$, is coterminal with $\displaystyle \small \frac{2\pi}{3}$.

We can figure this out by adding $\displaystyle \small 2\pi$, or equivalently $\displaystyle \small \frac{6\pi}{3}$ to get $\displaystyle \small \frac{2\pi}{3}$, or we can count thirds of pi around the unit circle clockwise. Either way, it is the only angle that ends in the second quadrant.

One way to uncover which quadrant this angle lies is to ask how many complete revolutions this angle makes by dividing it by 360 (and rounding down to the nearest whole number).

With a calculator we find that $\displaystyle 50000^o$ makes $\displaystyle 138$ full revolutions. Now the key lies in what the remainder the angle makes with $\displaystyle 138$ revolutions:

$\displaystyle 50000^o - 138 * 360^o = 320^o$

$\displaystyle 270^o < 320^o < 360^o$, therefore our angle lies in the fourth quadrant.

Alternatively, we could find evaluate $\displaystyle sin \ 50000^o$ and $\displaystyle cos \ 50000^o$.

The former (sine) gives us a negative number whereas the latter (cosine) gives a positive. The only quadrant in which sine is negative and cosine is positive is the fourth quadrant.

Step 1: Define the quadrants and the angles that go in:

QI:

$\displaystyle 0^\circ< x\leq 90^\circ$

QII:

$\displaystyle 90^\circ < x \leq 180^\circ$

QIII:

$\displaystyle 180^\circ < x \leq 270^\circ$

QIV:

$\displaystyle 270^\circ < x \leq 360$

Step 2: Find the quadrant where $\displaystyle 135^\circ$ is:

The angle is located in QII (Quadrant II)

First, using the unit circle, we can see that the denominator has a four in it, which means it is a multiple of $\displaystyle \pi/4$.

We want to reduce the angle down until we can visualize which quadrant it is in. You can subtract $\displaystyle 2\pi$ away from the angle each time (because that is just one revolution, and we end up at the same spot).

If you subtract away $\displaystyle 2\pi$ twice, you are left with $\displaystyle \pi/4$, which is in quadrant I. 

$\displaystyle 17\pi/4-8\pi/4-8\pi/4=\pi/4$.

The second quadrant contains angles between $\displaystyle \frac{\pi }{2}$ and $\displaystyle \pi$, plus those with additional multiples of $\displaystyle 2\pi$.  The angle $\displaystyle \frac{11\pi }{4}$ is, after subtracting $\displaystyle 2\pi$, is simply $\displaystyle \frac{3\pi }{4}$, which puts it in the second quadrant.

This problem relies on understanding reference angles and coterminal angles. A reference angle $\displaystyle R$ for an angle $\displaystyle \theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\displaystyle \theta$. A table of reference angles for each quadrant is given below.

Since $\displaystyle \cos \theta=-.3333$ is negative, solutions for $\displaystyle \theta$ will be in Quadrants II and III because these are the quadrants where cosine is negative.

Use inverse cosine and a calculator to find $\displaystyle \theta$:

$\displaystyle \cos R = -.3333$

$\displaystyle \cos^-^1(.3333)=70.53^\circ = R$

In Quadrant II, we have $\displaystyle R=180^\circ-\theta$, so $\displaystyle \theta=180^\circ-70.53^\circ=109.47^\circ$.

In Quadrant III, $\displaystyle R=\theta-180^\circ$, so $\displaystyle \theta=180^\circ+R=180^\circ+70.53^\circ=250.53^\circ$. 

Therefore $\displaystyle \theta=109.47^\circ$ and $\displaystyle 250.53^\circ$.

This problem relies on understanding reference angles and coterminal angles. A reference angle $\displaystyle R$ for an angle $\displaystyle \theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\displaystyle \theta$. A table of reference angles for each quadrant is given below.

Since $\displaystyle \tan\theta=-1.3456$ is negative, solutions for $\displaystyle \theta$ will be in Quadrants II and IV because these are the quadrants where tangent is negative. Use inverse tangent and a calculator to find $\displaystyle \theta$:

$\displaystyle \tan R=-1.3456$

$\displaystyle \tan^-^1(1.3456)=53.38^\circ=R$

In Quadrant II, we have $\displaystyle R=180^\circ-\theta$,  so $\displaystyle \theta=180^\circ-53.38^\circ=126.62^\circ$.

In Quadrant IV, $\displaystyle R=360^\circ-\theta$, so $\displaystyle R=360^\circ-53.38^\circ=306.62^\circ$.

Therefore $\displaystyle \theta=126.62^\circ$and $\displaystyle 306.62^\circ$.

We can use reference angles, inverse trig, and a calculator to solve this problem. Below is a table of reference angles. 

We have $\displaystyle \sin \theta=-.1919$ so $\displaystyle \sin^-^1(.1919)=11.06^\circ = R$. Next, think about where sine is negative, or reference the Function Signs column of the above table. Sine is negative in Quadrants III and IV.

In Quadrant III, $\displaystyle \theta=180^\circ+R=180^\circ+11.06^\circ=191.06^\circ$.

In Quadrant IV, $\displaystyle \theta=306^\circ-R=360^\circ-11.06^\circ=348.94^\circ$.

If this problem asked for values of $\displaystyle \theta$ between $\displaystyle 0^\circ$ and $\displaystyle 360^\circ$, our work would be done, but this problem does not restrict the range, so we need to give all possible values of $\displaystyle \theta$ by generalizing our answers. To do this, we must understand that all angles that are coterminal to $\displaystyle 191.06^\circ$ and $\displaystyle 348.94^\circ$ will also be solutions. Coterminal angles add or subtract multiples of $\displaystyle 360^\circ$. To write this generally, we write:

$\displaystyle \theta=191.06^\circ+n360^\circ$ and $\displaystyle \theta=348.94^\circ+n360^\circ$.