(14) Given that the coefficient of $x^{2}$ in the expansion of $(x \cos \theta+1)^{5}$ is equal to the coefficient of $x^{3}$ in the expansion of $\left(x+\frac{5}{4}\right)^{4}$, then $\cos \theta=$ $\qquad$
14 By the binomial theorem, the coefficient of $x^{2}$ in the expansion of $(x \cos \theta+1)^{5}$ is $\mathrm{C}_{5}^{3} \cdot \cos ^{2} \theta$, and the coefficient of $x^{3}$ in the expansion of $\left(x+\frac{5}{4}\right)^{4}$ is $\mathrm{C}_{4}^{1} \cdot \frac{5}{4}$. Therefore, we have $\mathrm{C}_{5}^{3} \cdot \cos ^{2} \theta=\mathrm{C}_{4}^{1} \cdot \frac{5}{4}$, solving this gives $\cos ^{2} \theta=\frac{1}{2}$, hence $\cos \theta= \pm \frac{\sqrt{2}}{2}$.