$-3 = (3\theta+\sin\theta) \cos(\theta)$ How do I solve this trig equation? How do I solve the following?
$$-3 = (3\theta+\sin\theta)( \cos\theta)$$
I'm supposed to get to $\theta =2.01692$ 
calculator use is ok 
EDIT:  I ended up just using a TI84 calculator and setting y1 = -3 and y2 = (.....) and finding the intersection points.
 A: As suggested in the comments, one approach is to use a root finding method, for example, Newton's Method.
We have:
$$f(\theta) = 3 + (3\theta+\sin\theta)( \cos\theta)$$
A plot shows that there is a root around $2$, so we will choose $\theta_0 = 2.5$.

The Newton-Raphson iteration is given by:
$$\theta_{n+1} = \theta_n - \dfrac{f(\theta_n)}{f'(\theta_n)} = \theta_n - \dfrac{(3 \theta_n+\sin (\theta_n)) \cos (\theta_n)+3}{\cos (\theta_n) (\cos (\theta_n)+3)-\sin (\theta_n) (3 \theta_n + \sin (\theta_n))}$$


*

*$\theta_0 = 2.5$

*$\theta_1 = 1.97217$

*$\theta_2 = 2.01714$

*$\theta_3 = 2.01692$

*$\theta_4 = 2.01692$


It only took four iterations to converge. You can repeat this using some of the other methods linked above and this can be programmed into most calculators that support such things.
I am not sure what your second statement means.
A: I had good luck with single point iteration.
$$ \theta \rightarrow \arccos\left(-\frac{1}{\theta + \frac{1}{3} \sin \theta} \right) $$
with just ten iterations you can reach the desired values, if you start from 
$\theta=1$.
Just rewrite the expression in terms of $cos \theta = -\frac{3}{3 \theta + \sin \theta}$ and keep evaluating it over and over again with the new values of $\theta$.
