jmc

Complete the square by adding 1 to each side. Then $(x+1{)}^2=1+i=e^{\frac{i\pi }{4}}\sqrt{2}$, so $x+1=\pm e^{\frac{i\pi }{8}}\sqrt[4]{2}$. The desired product is then $$\begin{array}{rl}\left(-1+\cos \left(\frac{\pi }{8}\right)\sqrt[4]{2}\right)\left(-1-\cos \left(\frac{\pi }{8}\right)\sqrt[4]{2}\right)& =1-{\cos }^2\left(\frac{\pi }{8}\right)\sqrt{2}\\ & =1-\frac{\left(1+\cos \left(\frac{\pi }{4}\right)\right)}{2}\sqrt{2}\\ & =\boxed{\frac{1-\sqrt{2}}{2}}.\end{array}$$