alors on aura
![{\displaystyle z^{3}+{\frac {1}{z^{3}}}={\frac {2}{\sin 2u}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f38cfee35d2c35ea2655d3f61cf0e052af7df53f)
d’où l’un tire
![{\displaystyle \sin 2u={\frac {1}{h}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1b19d3db4b9aedf501384e59b8e07f42c7202c8)
on aura donc
par les Tables des sinus. Si l’on fait ensuite
![{\displaystyle {\sqrt[{3}]{\operatorname {tang} u}}=\operatorname {tang} u',}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7dd4833df37a097f48e56a39ef8464d31fe47cb)
on aura
![{\displaystyle x={\frac {2{\sqrt {\frac {p}{3}}}}{\sin 2u'}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d79fb6f94f9c77f480f403e3657342870415d480)
C’est la valeur réelle de
ses deux valeurs imaginaires sont
![{\displaystyle x={\sqrt {\frac {p}{3}}}\left({\frac {-1\pm {\sqrt {-3}}\cos 2u'}{\sin 2u'}}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/313a5d3d0c0bfb47ea6fc41d6bdc5c6c5421369e)
Il nous reste à considérer le cas où l’on a
![{\displaystyle x={\sqrt {\frac {p}{3}}}\left(z-{\frac {1}{z}}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fe0ac86a834414fc3a2ab735523b48cc69caf1e)
On fera, dans ce cas,
et l’équation
![{\displaystyle z^{3}-{\frac {1}{z^{3}}}=2h}](https://wikimedia.org/api/rest_v1/media/math/render/svg/012ba4d213b0a428ef6be88e6e8e162da6074eaf)
donnera
![{\displaystyle \operatorname {tang} 2u={\frac {1}{h}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c69fa8b352f9f99769b4faaedfc27bf1d0fc4b42)
on aura donc l’angle
au moyen des Tables. Soit
![{\displaystyle \operatorname {tang} u'={\sqrt[{3}]{\operatorname {tang} u}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b193d1d1532d20ea7dde4e9b652887db28fd2bc5)
on aura
![{\displaystyle x={\frac {2{\sqrt {\cfrac {p}{3}}}}{\operatorname {tang} 2u'}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c26e449da0ed3cb261a2f65c00ce4243468480c9)
Les valeurs imaginaires de
sont
![{\displaystyle x={\sqrt {\frac {p}{3}}}\left({\frac {\cos 2u'\pm {\sqrt {-3}}}{\sin 2u'}}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/877f61f085e181fcbe5267fb4927048b8f90d019)