would like to solve the equation for theta:
z = b*cos(theta) + a*cos(theta)
been a long time since i took trig, any help would be appreciated.
Are you sure you didn't mean $\displaystyle z= a cos(\theta)+ b sin(\theta)$? That's a much more interesting question. As topsquark says, your equation is just a matter of combining those two "$\displaystyle cos(\theta)$ terms.
To solve $\displaystyle z= a cos(\theta)+ b sin(\theta)$, use the identity sin(A+B)= sin(A) cos(B)+ cos(A) sin(B). That would be the same as the previous equation if a= sin(A) and b= cos(A). The problem with that is that we would have to have $\displaystyle a^2+ b^2= sin^2(A)+ cos^2(A)$ which is not, in general, true. We can fix that factoring $\displaystyle \sqrt{a^2+ b^2}$ out. That is, we write the equation as
$\displaystyle z= \sqrt{a^2+ b^2}\left(\frac{a}{\sqrt{a^2+ b^2}}cos(\theta)+ \frac{b}{\sqrt{a^2+b^2}} sin(\theta)\right)$ so that
$\displaystyle \frac{z}{\sqrt{a^2+b^2}}= \left(\frac{a}{\sqrt{a^2+ b^2}}cos(\theta)+ \frac{b}{\sqrt{a^2+b^2}} sin(\theta)\right)$
So that now $\displaystyle \left(\frac{a}{\sqrt{a^2+ b^2}}\right)^2+ \left(\frac{b}{\sqrt{a^2+ b^2}\right)^2}= \frac{a^2+ b^2}{a^2+ b^2}= 1$
Now, we can let $\displaystyle \frac{a}{\sqrt{a^2+ b^2}}= sin(\phi)$ and $\displaystyle \frac{b}{\sqrt{a^2+b^2}}= cos(\phi)$ so that the equation can be written
$\displaystyle sin(\phi)cos(\theta)+ cos(\phi)sin(\theta)= sin(\theta+ \phi)= \frac{z}{\sqrt{a^2+ b^2}}$
then $\displaystyle \theta+ \phi= arcsin\left(\frac{z}{\sqrt{a^2+ b^2}}\right)$
$\displaystyle \theta= arcsin\left(\frac{z}{\sqrt{a^2+ b^2}}\right)- \phi= arcsin\left(\frac{z}{\sqrt{a^2+ b^2}}\right)- arcsin(\frac{a}{\sqrt{a^2+ b^2}})$