$$\quad \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { 6 } \theta = \frac { 5 } { 6 } \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { 4 } \theta d \theta$$
$\exists n_{2}\in \mathrm{Z}\text{ : }\theta =\frac{2\pi n_{2}-\arccos(\frac{2^{\frac{2}{3}}\left(\sqrt[3]{2}-\sqrt[3]{5}\right)}{2})+2\pi }{2}\text{ or }\exists n_{1}\in \mathrm{Z}\text{ : }\theta =\frac{2\pi n_{1}+\arccos(\frac{2^{\frac{2}{3}}\left(\sqrt[3]{2}-\sqrt[3]{5}\right)}{2})}{2}$
