The maximum value of the expression $\frac{1}{\sin^{2}\theta + 3\sin\theta\cos\theta + 5\cos^{2}\theta}$ is ______.

The maximum value of the expression 1/(sin²θ + 3sinθcosθ + 5cos²θ) is ______. $E=\dfrac{1}{\sin^2\theta+3\sin\theta\cos\theta+5\cos^2\theta}$ $D=\sin^2\theta+3\sin\theta\cos\theta+5\cos^2\theta$ $\text{Find: } \max E$ $\text{Since } D>0,\quad \max E=\dfrac{1}{\min D}$ $D=\dfrac{1-\cos 2\theta}{2}+3\cdot\dfrac{\sin 2\theta}{2}+5\cdot\dfrac{1+\cos 2\theta}{2}$ $D=\dfrac{6+4\cos 2\theta+3\sin 2\theta}{2}$ $=3+\dfrac{1}{2}(4\cos 2\theta+3\sin 2\theta)$ $4\cos 2\theta+3\sin 2\theta=5\cos(2\theta-\alpha),\quad \tan\alpha=\dfrac{3}{4}$ $-5\le 4\cos 2\theta+3\sin 2\theta\le 5$ $\text{amplitude}=\sqrt{4^2+3^2}=\sqrt{25}=5$ $D_{\min}=3+\dfrac{1}{2}(-5)=3-\dfrac{5}{2}=\dfrac{1}{2}$ $E_{\max}=\dfrac{1}{D_{\min}}=\dfrac{1}{1/2}=2$ $\boxed{E_{\max}=2}$ $\cos 2\theta=-\dfrac{4}{5},\ \sin 2\theta=-\dfrac{3}{5}\Rightarrow 4\cos 2\theta+3\sin 2\theta=-5$ $D=3-\dfrac{5}{2}=\dfrac{1}{2}\ \Rightarrow\ E=\dfrac{1}{1/2}=2\ \checkmark$