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

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

I-solve ang b

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

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

Mga Hakbang sa Pagso-solve ng Mga Radical Equation

Pagpalitin ang magkabilang panig para nasa kaliwang bahagi ang lahat ng variable na term.

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

Gamitin ang distributive property para i-multiply ang \sqrt{b} gamit ang 1-\left(\sin(a)\right)^{2}.

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

Pagsamahin ang lahat ng term na naglalaman ng b.

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

I-divide ang magkabilang dulo ng equation gamit ang 1-\left(\sin(a)\right)^{2}.

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

Kapag na-divide gamit ang 1-\left(\sin(a)\right)^{2}, ma-a-undo ang multiplication gamit ang 1-\left(\sin(a)\right)^{2}.

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

I-divide ang \sin(a) gamit ang 1-\left(\sin(a)\right)^{2}.

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

I-square ang magkabilang dulo ng equation.

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

I-solve ang a

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