3tanx=5K eching

$$3 \tan x = 5 K$$

K uchun yechish

$$K=\frac{3\tan(x)}{5}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$5K=3\tan(x)$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5K}{5}=\frac{3\tan(x)}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$K=\frac{3\tan(x)}{5}$$

x uchun yechish

$$x=2\pi n_{1}+\arcsin(\frac{5K}{\sqrt{25K^{2}+9}})+\pi \text{, }n_{1}\in \mathrm{Z}\text{, }\exists n_{3}\in \mathrm{Z}\text{ : }\left(n_{1}>\frac{2n_{3}-\frac{2\arcsin(\frac{5K}{\sqrt{25K^{2}+9}})}{\pi }-1}{4}\text{ and }n_{1}<\frac{2n_{3}-\frac{2\arcsin(\frac{5K}{\sqrt{25K^{2}+9}})}{\pi }+1}{4}\right)$$

$$x=2\pi n_{2}+\arcsin(\frac{5K}{\sqrt{25K^{2}+9}})\text{, }n_{2}\in \mathrm{Z}\text{, }\exists n_{3}\in \mathrm{Z}\text{ : }\left(n_{3}>\frac{4n_{2}+\frac{2\arcsin(\frac{5K}{\sqrt{25K^{2}+9}})}{\pi }-3}{2}\text{ and }n_{3}<\frac{4n_{2}+\frac{2\arcsin(\frac{5K}{\sqrt{25K^{2}+9}})}{\pi }-1}{2}\right)$$