Solve for $x$ in $(3sf)$, where $\cos(x) - \tan(x) = 3 $. The problem I am struggling with to solve is this. I have already tried to solve it however I end up with quadratic $\sin$ or $\tan$ whichever way you do it, which does not help. 
Solve for $x$ $(3sf)$ where:
$\cos(x) - \tan(x) = 3 $
from $0$ to $2\pi$

Let $$f(x)=\cos(x)-\tan(x)-3$$
$$f'(x)=\sin(x)-\frac{1}{\cos^2(x)}$$
Now apply Newton method, by


*

*$x_0=2$ (guess something not too far by looking at a graph of the function $f$)

*$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$


And $x_{25}=1.865810565186072908108$ which is a very good approximation.
If you start with $x_0=5$, you'll find the second solution. 
