1. (2004 National High School Mathematics League Sichuan Preliminary) Given the inequality $m^{2}+\left(\cos ^{2} \theta-5\right) m+4 \sin ^{2} \theta \geqslant 0$ always holds, find the range of the real number $m$.
1. The original expression is equivalent to $m^{2}+m \cos ^{2} \theta-5 m+4-4 \cos ^{2} \theta \geqslant 0$, which simplifies to $(m-4) \cos ^{2} \theta+m^{2}-5 m+4 \geqslant 0$. Let $\cos ^{2} \theta=$ $t \in[0,1], f(t)=(m-4) t+m^{2}-5 m+4 \geqslant 0$, then $\left\{\begin{array}{l}f(0)=m^{2}-5 m+4 \geqslant 0 \\ f(1)=m^{2}-4 m \geqslant 0\end{array} \Rightarrow m \geqslant 4\right.$ or $m \leqslant 0$.