Solving nested trigonometric function with additional linear function I have a graph as follows. How would I go about solving it? The "x" is a variable not multiplication. I am studying senior maths C but am unable to solve it accurately (graphing is not accurate enough). Any assistance greatly appreciated.
$$20 = \cos(\tan(0.22x + 1.65)) + 1.5x$$
edit
It has multiple answers, all I require is the first.
 A: Best method I see for that is to solve using Newton's Method, which is a basic numerical root-finding algorithm. Convert your equation to an implicit homogeneous form $f(x)=\cos(\tan(0.22x+1.65))+1.5x-20=0$, and then apply Newton's Method, which is as follows:
$$x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}$$
This method works by obtaining the slope of the function at a given point, and then linearly extrapolating it back to the estimated zero. The process is repeated until the value of $x$ doesn't change to within some relative error.
$$relerr=\frac{x_{i}-x_{i-1}}{x_i}<=\epsilon$$
To start with, make some kind of guess as to the solution. Try a value near to where you want the solution to be, so if you want the first solution larger than zero, use $x_0=0$ as your guess.
Source:
http://en.wikipedia.org/wiki/Newton's_method
A: This is not answer but it is too long for a comment.
As FundThmCalculus answered, the simplest way would be Newton method but the problem is the starting point; depending on what could be selected, the iterative process could lead to any solution.
What is interesting is to notice that the function is clearly bounded since $-1 \leq \cos(\theta)\leq 1$ which implies that $$1.5x-21\leq\cos(\tan(0.22x+1.65))+1.5x-20\leq 1.5x-19$$ which implies $$\frac{38}{3}\leq x\leq 14$$ Even if this range can look quite small, it is not since it contains almost all the roots.
So, what I suggest, if we are only concerned by the first root, is to inspect the function using rather small steps (say $\Delta x =0.1$) or to plot the function between these two bounds. Doing so, it is clear that the first root is close to $13$. 
So, start Newton method using $x_0=13$; it will then generate the following iterates : $13.0494$, $13.0488$ which is the solution for six significant figures.
It must be noticed that, if instead, we had started using $x_0=\frac{38}{3}$, the iterative process would have generated : $13.3964$, $13.3362$, $13.3459$, $13.3456$ which corresponds to the second root. 
This shows how, with no care, we can easily miss the root we are looking for.
Edit
Assuming that we are not lazy, we could expand the function as a Taylor series built at $x=\frac{38}{3}$; limiting to second order, this would give $$f(x)\approx 4.9535 (x-\frac{38}{3})^2+2.63639 (x-\frac{38}{3})-1.92381=0$$ Solving the quadratic for the closest root leads to $x\approx 13.0782$ (not too bad but really tedious).
