Réitigh r=Q(1-sinθ)

$$r = Q ( 1 - \sin \theta )$$

Réitigh do Q.

$$\left\{\begin{matrix}Q=-\frac{r}{\sin(\theta )-1}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =2\pi n_{1}+\frac{\pi }{2}\\Q\in \mathrm{R}\text{, }&r=0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\theta =2\pi n_{1}+\frac{\pi }{2}\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$Q\left(1-\sin(\theta )\right)=r$$

Úsáid an t-airí dáileach chun Q a mhéadú faoi 1-\sin(\theta ).

$$Q-Q\sin(\theta )=r$$

Comhcheangail na téarmaí ar fad ina bhfuil Q.

$$\left(1-\sin(\theta )\right)Q=r$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(-\sin(\theta )+1\right)Q=r$$

Roinn an dá thaobh faoi 1-\sin(\theta ).

$$\frac{\left(-\sin(\theta )+1\right)Q}{-\sin(\theta )+1}=\frac{r}{-\sin(\theta )+1}$$

Má roinntear é faoi 1-\sin(\theta ) cuirtear an iolrúchán faoi 1-\sin(\theta ) ar ceal.

$$Q=\frac{r}{-\sin(\theta )+1}$$

Réitigh do r.

$$r=Q\left(-\sin(\theta )+1\right)$$