Problem (Algebra):
Let $x$ be a real number in the interval $[0,360]$ such that the four expressions $\sin x^{\circ}, \cos x^{\circ}$, $\tan x^{\circ}, \cot x^{\circ}$ take on exactly three distinct (finite) real values. Compute the sum of all possible values of $x$.

Solution:
If the four expressions take on three different values, exactly two of the expressions must have equal values. There are $\left(\begin{array}{l}4 \\ 2\end{array}\right)=6$ cases to consider:

Case 1: $\sin x^{\circ}=\cos x^{\circ}$ : Then $\tan x^{\circ}=\cot x^{\circ}=1$, violating the condition that there be three distinct values.

Case 2: $\sin x^{\circ}=\tan x^{\circ}$ : Because $\tan x^{\circ}=\frac{\sin x^{\circ}}{\cos x^{\circ}}$, either $\cos x^{\circ}=1$ or $\sin x^{\circ}=0$. However, in both of these cases, $\cot x^{\circ}$ is undefined, so it does not have a real value.

Case 3: $\sin x^{\circ}=\cot x^{\circ}$ : Then $\sin x^{\circ}=\frac{\cos x^{\circ}}{\sin x^{\circ}}$, and so $\sin ^{2} x^{\circ}=\cos x^{\circ}$. Rewrite using the Pythagorean identity to obtain $\cos ^{2} x^{\circ}+\cos x^{\circ}-1=0$, so $\cos x^{\circ}=\frac{-1+\sqrt{5}}{2}$ (the other root is outside the range of $\cos )$. Because $\cos x^{\circ}>0$, this equation has two solutions in $[0,360]$ : an angle $x_{0}^{\circ}$ in the first quadrant and the angle $\left(360-x_{0}\right)^{\circ}$ in the fourth quadrant. The sum of these two values is 360 .

Case 4: $\cos x^{\circ}=\tan x^{\circ}$ : Use similar logic as in the previous case to obtain the equation $\sin ^{2} x^{\circ}+$ $\sin x^{\circ}-1=0$, so now $\sin x^{\circ}=\frac{-1+\sqrt{5}}{2}$. Because $\sin x^{\circ}>0$, this equation has two solutions, one an angle $x_{0}^{\circ}$ in the first quadrant, and the other its supplement $\left(180-x_{0}\right)^{\circ}$ in the second quadrant. The sum of these two values is 180 .

Case 5: $\cos x^{\circ}=\cot x^{\circ}$ : In this case, $\tan x^{\circ}$ is undefined for reasons analogous to those in Case 2.

Case 6: $\tan x^{\circ}=\cot x^{\circ}$ : Thus $\tan ^{2} x^{\circ}=1$, hence $\tan x^{\circ}= \pm 1$. If $\tan x^{\circ}=1$, then $\sin x^{\circ}=\cos x^{\circ}$, which yields only two distinct values. So $\tan x^{\circ}=-1$, which occurs at $x=135$ and $x=315$. The sum of these values is 450 .

The answer is $360+180+450=\mathbf{9 9 0}$.

Consider the graphs of all four functions; notice first that 0, 90, 180, 270 are not solutions because either $\tan x^{\circ}$ or $\cot x^{\circ}$ is undefined at each value.





Start in the first quadrant. Let $x_{1}$ and $x_{2}$ be the values of $x$ such that $\cos x^{\circ}=\tan x^{\circ}$ and $\sin x^{\circ}=\cot ^{\circ}$, respectively, labeled $A$ and $B$ in the diagram. Because $\cos x^{\circ}=\sin (90-x)^{\circ}$ and $\cot x^{\circ}=\tan (90-x)^{\circ}, x_{1}+x_{2}=90$. One can also see that the graphs of $y=\cot x^{\circ}$ and $y=\tan x^{\circ} \operatorname{cross}$ at $x=45$, but so do the graphs of $y=\sin x^{\circ}$ and $y=\cos x^{\circ}$. So at $x=45$, there are only two distinct values, not three.



In the second quadrant, $\tan x^{\circ}=\cot x^{\circ}$ when $x=135$. Also, because $\tan x^{\circ}$ increases from $-\infty$ to 0 while $\cos x^{\circ}$ decreases from 0 to -1 , there exists a number $x_{3}$ such that $\tan x_{3}^{\circ}=\cos x_{3}^{\circ}$ (marked point $C$ in the diagram above).





In the third quadrant, $\tan x^{\circ}$ and $\cot x^{\circ}$ are positive, while $\sin x^{\circ}$ and $\cos x^{\circ}$ are negative; the only place where graphs cross is at $x=225$, but this value is not a solution because the four trigonometric functions have only two distinct values.



In the fourth quadrant, $\tan x^{\circ}=\cot x^{\circ}=-1$ when $x=315$. Because $\sin x^{\circ}$ is increasing from -1 to 0 while $\cot x^{\circ}$ is decreasing from 0 to $-\infty$, there exists a number $x_{4}$ such that $\sin x_{4}^{\circ}=\cot x_{4}^{\circ}$ (marked $D$ in the diagram above). Because $\cos x^{\circ}=\sin (90-x)^{\circ}=\sin (450-x)^{\circ}$ and $\cot x^{\circ}=\tan (90-x)^{\circ}=\tan (450-x)^{\circ}$, the values $x_{3}$ and $x_{4}$ are symmetrical around $x=225$, that is, $x_{3}+x_{4}=450$.

The sum is $\left(x_{1}+x_{2}\right)+(135+315)+\left(x_{3}+x_{4}\right)=90+450+450=\mathbf{9 9 0}$.

Final answer: 990