y=sin(pi/3-2x)dx eching

$$y = \sin ( \frac { \pi } { 3 } - 2 x ) d x$$

d uchun yechish

$$\left\{\begin{matrix}d=\frac{y}{x\sin(\frac{\pi -6x}{3})}\text{, }&x\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{6}\\d\in \mathrm{R}\text{, }&\left(\exists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{6}\text{ or }x=0\right)\text{ and }y=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\sin(\frac{\pi }{3}-2x)dx=y$$

Tenglama standart shaklda.

$$x\sin(-2x+\frac{\pi }{3})d=y$$

Ikki tarafini \sin(\frac{1}{3}\pi -2x)x ga bo‘ling.

$$\frac{x\sin(-2x+\frac{\pi }{3})d}{x\sin(-2x+\frac{\pi }{3})}=\frac{y}{x\sin(-2x+\frac{\pi }{3})}$$

\sin(\frac{1}{3}\pi -2x)x ga bo'lish \sin(\frac{1}{3}\pi -2x)x ga ko'paytirishni bekor qiladi.

$$d=\frac{y}{x\sin(-2x+\frac{\pi }{3})}$$

y ni \sin(\frac{1}{3}\pi -2x)x ga bo'lish.

$$d=\frac{y}{x\sin(\frac{\pi -6x}{3})}$$