\frac{\cos 3x}{\cos 4x}=\left(1-\frac{9x^2}{2}+o(x^2)\right)
\left(1-\frac{16x^2}{2}+o(x^2)\right)^{-1}=\left(1-\frac{9x^2}{2}+o(x^2)\right)
\left(1+\frac{16x^2}{2}+o(x^2)\right)=\left(1+\frac{7x^2}{2}+o(x^2)\right)