Show that $|2\cdot\ln \cos x + x^2|\leq \frac{\pi^3}{48}$ for $|x|\leq\frac{\pi}{4}$ using Taylor's theorem.
My proof: Let $f(x)=\ln \cos x$ and write down the second-degree Taylor polynomial of $f$ at $0$ which is $$T_0(x)=-\frac{x^2}{2}$$
Using Taylor's theorem and calculating $f'''(0)=0$ we get that $$\forall x\in(-\pi/2,\pi/2):f(x)=-\frac{x^2}{2}+0\cdot f'''(\eta_x)$$ for some $\eta_x$ with $|\eta_x-0|\leq|x-0|$.
Hence: $$|2\cdot\ln \cos x + x^2|=0\;\forall x\in(-\pi/2,\pi/2)\;\;(1)$$
which is very unlikely since we are asked to show the statement for $|x|\leq \frac{\pi}{4}$ but according to what I have it holds for arbitrary $|x|< \frac{\pi}{2}$ and $(1)$ also seems very unlikely to be true.
Because you want just to the second degree, you can calculate the polynomial directly. You have $$ f'(x)=-\frac{\sin x}{\cos x},\ \ f''(x)=-\frac1{\cos^2x},\ \ f'''(x)=-\frac{2\sin x}{\cos^3x}. $$ As $$f(0)=0,\ \ f'(0)=0,\ \ f''(0)=-1.$$ This gives you have $$ \ln(\cos x)=-\frac{x^2}2-\frac{2\sin \xi}{\cos^3\xi}\,\frac{x^3}6 $$ for some $\xi$ between $0$ and $x$. So $$ |2f(x)+x^2|=\frac{2|\sin\xi|}{|\cos\xi|^3}\,\frac{x^3}6. $$ Because $|x|\leq\tfrac\pi4$, you have $|\xi|\leq\tfrac\pi4$ and so $$ \cos\xi\geq\frac1{\sqrt2},\ \ |\sin\xi|\leq\frac1{\sqrt2}. $$ Then, when $|x|\leq\tfrac\pi4$, $$ |2f(x)+x^2|\leq\frac{2/\sqrt2}{(1/\sqrt2)^3}\,\frac{(\pi/4)^3}6=\frac46\,\frac{\pi^3}{4^3}=\frac{\pi^3}{96}<\frac {\pi^3}{48}. $$