f=a(1+cosθ) eching

$$f = a ( 1 + \cos \theta )$$

a uchun yechish

$$\left\{\begin{matrix}a=\frac{f}{\cos(\theta )+1}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =2\pi n_{1}+\pi \\a\in \mathrm{R}\text{, }&f=0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\theta =2\pi n_{1}+\pi \end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

a ga 1+\cos(\theta ) ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$f=a+a\cos(\theta )$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$a+a\cos(\theta )=f$$

a'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(1+\cos(\theta )\right)a=f$$

Tenglama standart shaklda.

$$\left(\cos(\theta )+1\right)a=f$$

Ikki tarafini 1+\cos(\theta ) ga bo‘ling.

$$\frac{\left(\cos(\theta )+1\right)a}{\cos(\theta )+1}=\frac{f}{\cos(\theta )+1}$$

1+\cos(\theta ) ga bo'lish 1+\cos(\theta ) ga ko'paytirishni bekor qiladi.

$$a=\frac{f}{\cos(\theta )+1}$$

f uchun yechish

$$f=a\left(\cos(\theta )+1\right)$$