Réitigh sina=sqrt{b}(1-sin^2a)

$$\sin a = \sqrt { b } ( 1 - \sin ^ { 2 } a )$$

Réitigh do b.

$$b=\frac{\left(\tan(a)\right)^{2}}{\left(\cos(a)\right)^{2}}$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }\left(\left(a\geq 2\pi n_{1}\text{ and }a<2\pi n_{1}+\frac{\pi }{2}\right)\text{ or }\left(a>2\pi n_{1}+\frac{\pi }{2}\text{ and }a\leq 2\pi n_{1}+\pi \right)\right)$$

Céimeanna chun cothromóidí fréimhe a réiteach

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

$$\sqrt{b}\left(1-\left(\sin(a)\right)^{2}\right)=\sin(a)$$

Úsáid an t-airí dáileach chun \sqrt{b} a mhéadú faoi 1-\left(\sin(a)\right)^{2}.

$$\sqrt{b}-\sqrt{b}\left(\sin(a)\right)^{2}=\sin(a)$$

Comhcheangail na téarmaí ar fad ina bhfuil b.

$$\left(1-\left(\sin(a)\right)^{2}\right)\sqrt{b}=\sin(a)$$

Roinn an dá thaobh faoi 1-\left(\sin(a)\right)^{2}.

$$\frac{\left(-\left(\sin(a)\right)^{2}+1\right)\sqrt{b}}{-\left(\sin(a)\right)^{2}+1}=\frac{\sin(a)}{-\left(\sin(a)\right)^{2}+1}$$

Má roinntear é faoi 1-\left(\sin(a)\right)^{2} cuirtear an iolrúchán faoi 1-\left(\sin(a)\right)^{2} ar ceal.

$$\sqrt{b}=\frac{\sin(a)}{-\left(\sin(a)\right)^{2}+1}$$

Roinn \sin(a) faoi 1-\left(\sin(a)\right)^{2}.

$$\sqrt{b}=\frac{\tan(a)}{\cos(a)}$$

Cearnaigh an dá thaobh den chothromóid.

$$b=\frac{\left(\tan(a)\right)^{2}}{\left(\cos(a)\right)^{2}}$$

Réitigh do a.

$$\left\{\begin{matrix}a=2\pi n_{2}+\arcsin(\frac{\sqrt{4b+1}-1}{2\sqrt{b}})\text{, }n_{2}\in \mathrm{Z}\text{; }a=2\pi n_{3}-\arcsin(\frac{\sqrt{4b+1}-1}{2\sqrt{b}})+\pi \text{, }n_{3}\in \mathrm{Z}\text{, }&b>0\\a=\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&b=0\end{matrix}\right.$$