Solve sina=sqrt{b}(1-sin^2a)

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

Solve for b

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

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

Steps for Solving Radical Equations

Swap sides so that all variable terms are on the left hand side.

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

Use the distributive property to multiply \sqrt{b} by 1-\left(\sin(a)\right)^{2}.

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

Combine all terms containing b.

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

Divide both sides by 1-\left(\sin(a)\right)^{2}.

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

Dividing by 1-\left(\sin(a)\right)^{2} undoes the multiplication by 1-\left(\sin(a)\right)^{2}.

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

Divide \sin(a) by 1-\left(\sin(a)\right)^{2}.

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

Square both sides of the equation.

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

Solve for a

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