배각공식 방정식

$\displaystyle 3 ( 1 - 2 \sin ^{2} \theta ) + 4 \sqrt{2} \sin \theta - 3 = 0$ $\displaystyle 6 \sin ^{2} \theta - 4 \sqrt{2} \sin \theta = 0$ $\displaystyle \sin \theta ( 6 \sin \theta - 4 \sqrt{2} ) = 0$ $\displaystyle 0 < \theta < \frac{\pi}{2}$이므로 $\sin \theta \neq 0$ ∴ $\displaystyle \sin \theta = \frac{2 \sqrt{2}}{3}$ $\displaystyle \cos ^{2} \theta = 1 - \sin ^{2} \theta = 1 - \frac{8}{9} = \frac{1}{9}$ $\displaystyle 0 < \theta < \frac{\pi}{2}$에서 $\cos \theta > 0$이므로 $\displaystyle \cos \theta = \frac{1}{3}$