solve for $\theta$ in $ \theta - \sin \theta = 0.6\pi $ I have simplified my equation to the following. I can't quite remember how to solve for $\theta$. 
$$ \theta - \sin \theta = 0.6\pi $$
I tried to rearrange and make use of $\sin^2\theta + \cos^2\theta = 1$, but I couldn't quite isolate $1-\sin^2\theta$. I got as far as:
$$1.2\pi+0.36\pi^2=1-\frac{\sin^2\theta}{\theta^2}$$
I'm not sure what else to try. Any suggestions?
 A: As Robert Israel already answered, there is no closed form solution to equations which mix polynomial and trigonometric terms (this is already the case for $x=\cos(x)$) and numerical methods are required.
Methods such as Newton require a "resonable" starting value what you could obtain here at the price of a very crude approximation such as $$\sin(\theta)\approx \frac{120 (\pi -\theta) \theta}{\pi ^5}$$ (for $0 \leq \theta \leq \pi)$. Replacing in the equation and solving the quadratic gives as an approximate solution $$\theta=\frac{600 \pi -5 \pi ^5+\sqrt{360000 \pi ^2+1200 \pi ^6+25 \pi ^{10}}}{1200}\approx 2.50804$$ Using this result as a starting point, let us use Newton iterative scheme which, starting from the guess $\theta_0$, will update it according to $$\theta_{n+1}=\theta_n-\frac{f(\theta_n)}{f'(\theta_n)}$$ So, using $$f(\theta)=\theta-\sin(\theta)-\frac 3 5 \pi$$ $$f'(\theta)=1-\cos(\theta)$$ the following iterates will be obtained $$\theta_0=2.50804$$ $$\theta_1=2.49083$$ $$\theta_2=2.49078$$ which is the solution for six significant figures.
Similarly, if, from the plot of the function, you notice that the solution is just above $\frac {3\pi}4$, you could expand $f(\theta)$ as a Taylor series around this point and get $$f(\theta)=\left(\frac{3 \pi
   }{20}-\frac{1}{\sqrt{2}}\right)+\left(1+\frac{1}{\sqrt{2}}\right)
   \left(\theta-\frac{3 \pi }{4}\right)+O\left(\left(\theta-\frac{3 \pi
   }{4}\right)^2\right)$$ Ignoring the high order terms, this will give as an approximation $$\theta=-1+\sqrt{2}+\frac{3}{20} \left(3+\sqrt{2}\right) \pi \approx 2.49436$$
A: There is no "closed-form" solution.  You can use numerical methods, such as Newton's method, to find an approximate solution.
