Expand the function in a Maclaurin series $\ln(5\cos^{3}(x))$ $$\ln(5\cos^{3}(x))$$ Need to be expanded: $x^{4}$
I need to end this problem.
So I laid the beginning of the function.
$$\cos x=1-\frac{x^2}{2!}-\frac{x^4}{4!}+o(x^4)$$
$$5\cos^3(x)=5\left(1-\frac{x^2}{2!}+o(x^4)\right)^3=5-\frac{15x^2}{2!}+\frac{35}{8}x^4+o(x^4).$$
I do not know what to do next.

If you want to be cute about it, note that $\cos x = \sqrt{1 - \sin^2 x}$, so
$$
\ln(5\cos^3 x)=\ln 5 + 3\ln\cos x=\ln 5 + \frac{3}{2}\ln(1-\sin^2 x)=\ln 5 - \frac{3}{2}\sin^2 x - \frac{3}{4}\sin^4 x + O(x^6).
$$
To get the $x^4$ term, you just need $\sin^2 x \approx (x - x^3/6)^2 \approx x^2 - x^4/3$, so
$$
\ln(5\cos^3 x)\approx \ln 5 - \frac{3}{2}\left(x^2-\frac{1}{3}x^4\right)-\frac{3}{4}x^4=\ln 5 - \frac{3}{2}x^2 - \frac{1}{4}x^4.
$$
