To find a real root find real root of given equation
$$5x - 2 \cos x -1=0 $$
I only know that I should use 
$$x_n = 
\frac{ x_{n-2} f_{n-1} - x_{n-1} f_{n-2} } { f_{n-1} - f_{n-2} }$$
And I applied it, but didn't get desired answer
Can anyone explain me about some important point, to solve this type of problems.
 A: As $f'(x)=5+2\sin(x)$ has no root, the function is monotonic and has at most one root.
Given the range of the cosine, you can write the bracketing
$$5x-3\le5x-2\cos x-1\le5x+1,$$ which gives two simple estimates, and $x\in(-\frac15,\frac35)$, from which you can start regula falsi.

A: Using Ian's comment and Yves Daoust's answer, we get an initial value in the interval $(-1/5,3/5)$.
Since there is only one real root it makes sense to take an initial value inside this interval, for example $x_0=1/2$.
I don't see why you need a complicated procedure. The easiest way (for me) appears to be fixed point iteration:
$$x_n=\frac{1+2 \cos x_{n-1}}{5}$$
$$x_0=\frac{1}{2}=0.5$$
$$x_1=\color{blue}{0.5}51033024756149$$
$$x_2=\color{blue}{0.54}0793647370719$$
$$x_3=\color{blue}{0.542}920147441434$$
$$x_4=\color{blue}{0.542}481469458663$$
$$x_5=\color{blue}{0.5425}72091664505$$
The root is:
$$x=\color{blue}{0.5425565799894813}$$

If you need faster convergence, you can of course use another method.
A: By convexity there is just one real solution of $\cos(x)=\frac{5x-1}{2}$, and you may apply Newton's method with starting point $x_0=\frac{\pi}{6}$ to get it is $\approx 0.54255658$ with very few iterations.
