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

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

b uchun yechish

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

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

Ildizli tenglamalarni yechish qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

\sqrt{b} ga 1-\left(\sin(a)\right)^{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

b'ga ega bo'lgan barcha shartlarni birlashtirish.

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

Ikki tarafini 1-\left(\sin(a)\right)^{2} ga bo‘ling.

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

1-\left(\sin(a)\right)^{2} ga bo'lish 1-\left(\sin(a)\right)^{2} ga ko'paytirishni bekor qiladi.

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

\sin(a) ni 1-\left(\sin(a)\right)^{2} ga bo'lish.

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

a uchun yechish

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