Semi-infinite plate problem when 2 edges are equal temperature and one edge is a function of position Here is the problem posted:

Now here is my solution for a)

I knew to discard the $e^{ky}$ and $\sin(kx)$ due to $T\to 20$ when $y\to\infty$ and $T= 20$ at $x = 0$.
Which leaves me with $\cos(kx)*e^{-ky} = T$.
Now I have two main problems. 
1) I don't know how to solve for $k$.
If I plug in $(0,0)$ I get $cos(0*k) = 20$
2) I don't know if I'm understanding the problem correctly, because if I plug in $x = 0$ to the bottom edge equation, I get $T = 20$ which equals the given left edge temperature.
But if I plug in $x = 30$ I get $T = 320$, not 20, which should be the right edge.
I really need help on this.

I suggest you start making $t=T-20$ your new temperature function; recall that the Laplace operator is linear, and if $T$ is a solution so is $t$ and vice versa. The advantage of this new temperature is that it makes 3 out of 4 boundary conditions homogeneous, making the problem easier to handle.
More precisely, if you set $t=T-20$ then $t(x=0)=t(x=30)=t(x,y=\infty)=0$. 
Now let $t=f(x)g(y)$, separation of variables yields:
$$f=-k^2f_{xx} \text{ and } g=k^2g_{yy}.$$
Making $f(x)=A\sin{kx}+B\cos{kx}$ and using the boundary conditions at $x=0$ and $x=30$ we get $B=0$ and $k=\frac{n\pi}{30}, n\in\{1,2,3,\dots\}$. 
The solutions for $f$ have the form:
$$f_n=A_n \sin{\left(\frac{n\pi}{30}x\right)}$$
Proceeding in an analogous way for $g$ we get
$g(y)=Ce^{ky}+De^{-ky}$ and from $t\rightarrow 0$ as $y\rightarrow \infty$ we deduce $C=0$, so, taking into account the previously calculated values for $k$:
$$g_n=D_n e^{-\frac{n\pi}{30}y}.$$
Now note that each $t_n=f_n g_n$ is a solution, and so is their sum; the solution can be therefore written as:
$$t(x,y)=\sum_{n=1}^\infty A_n \sin{\left(\frac{n\pi}{30}x\right)}\exp{\left(-\frac{n\pi}{30}y\right)}.$$
Where I have renamed the constants. We have only to determine the coefficients $A_n$, we use the condition at $y=0$ for this. From $t(x,y=0)=10x$ we obtain
$$10x=\sum_{n=1}^\infty A_n \sin{\left(\frac{n\pi}{30}x\right)}.$$
Multiplying by $\sin{\left( \frac{m\pi}{30}x\right)}$, changing variables, integrating and using the orthogonality relations between sines we finally arrive at
$$A_n=10\frac{2}{\pi}\frac{\pi}{30}\int_0^{30}x \sin{\left( \frac{n\pi}{30}x \right)} \operatorname d\!x .$$
This is the Fourier series of your lower boundary conditions (I suggest you check this calculation as I might have made mistakes).
Once you calculate the coefficients you can go back to $T$ by summing the baseline 20 to $t$, the constant term of the Fourier series for the solution..
