\(\displaystyle 8\sin^2\theta +\sqrt{2}\cos \theta =7\)
\(\displaystyle 8\left(1-\cos^2\theta\right) +\sqrt{2}\cos \theta =7\)
\(8\cos^2\theta-\sqrt{2}\cos\theta-1=0\)
令 \(t=\cos\theta\)，則 \(8t^2-\sqrt{2} t-1=0\) 的兩根為 \(\cos\theta_1\) 與 \(\cos\theta_2\)
由根與係數關係式，可知 \(\displaystyle\cos\theta_1+\cos\theta_2=\frac{\sqrt{2}}{8}\) 且 \(\displaystyle\cos\theta_1 \cos\theta_2=\frac{-1}{8}\)
所求 \(\displaystyle=32\cos^2\frac{\theta _1}{2} \cos^2\frac{\theta _2}{2}=32\left(\frac{1+\cos\theta_1}{2}\right)\left(\frac{1+\cos\theta_2}{2}\right)=.........\)
-----------------如後續 thepiano 老師的回覆，我上面寫的掉入陷阱了～ＸＤ
因為 \(\theta_1,\theta_2\) 在第一、四象限的關係，所以 \(\displaystyle\cos\theta_1=\cos\theta_2 = \frac{\sqrt{2}+\sqrt{34}}{16}\)
