a-y/ax^2+y=cos^2x eching

$$\frac { a - y } { a x ^ { 2 } } + y = \cos ^ { 2 } x$$

a uchun yechish (complex solution)

$$\left\{\begin{matrix}a=\frac{2y}{-x^{2}\cos(2x)+2yx^{2}-x^{2}+2}\text{, }&y\neq 0\text{ and }y\neq \frac{\cos(2x)+1-\frac{2}{x^{2}}}{2}\text{ and }x\neq 0\\a\neq 0\text{, }&y=0\text{ and }\frac{\cos(2x)-\frac{2}{x^{2}}}{2}=-\frac{1}{2}\text{ and }x\neq 0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

a qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini ax^{2} ga ko'paytirish.

$$a-y+ax^{2}y=ax^{2}\left(\cos(x)\right)^{2}$$

Ikkala tarafdan ax^{2}\left(\cos(x)\right)^{2} ni ayirish.

$$a-y+ax^{2}y-ax^{2}\left(\cos(x)\right)^{2}=0$$

y ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$a+ax^{2}y-ax^{2}\left(\cos(x)\right)^{2}=y$$

a'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(1+x^{2}y-x^{2}\left(\cos(x)\right)^{2}\right)a=y$$

Tenglama standart shaklda.

$$\left(-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1\right)a=y$$

Ikki tarafini 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} ga bo‘ling.

$$\frac{\left(-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1\right)a}{-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1}=\frac{y}{-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1}$$

1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} ga bo'lish 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} ga ko'paytirishni bekor qiladi.

$$a=\frac{y}{-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1}$$

y ni 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} ga bo'lish.

$$a=\frac{y}{x^{2}\left(-\left(\cos(x)\right)^{2}+y\right)+1}$$

a qiymati 0 teng bo‘lmaydi.

$$a=\frac{y}{x^{2}\left(-\left(\cos(x)\right)^{2}+y\right)+1}\text{, }a\neq 0$$

a uchun yechish

$$\left\{\begin{matrix}a=\frac{y}{-\left(x\cos(x)\right)^{2}+yx^{2}+1}\text{, }&y\neq 0\text{ and }y\neq \left(\cos(x)\right)^{2}-\frac{1}{x^{2}}\text{ and }x\neq 0\\a\neq 0\text{, }&y=0\text{ and }0=\left(\cos(x)\right)^{2}-\frac{1}{x^{2}}\text{ and }x\neq 0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

a qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini ax^{2} ga ko'paytirish.

$$a-y+ax^{2}y=ax^{2}\left(\cos(x)\right)^{2}$$

Ikkala tarafdan ax^{2}\left(\cos(x)\right)^{2} ni ayirish.

$$a-y+ax^{2}y-ax^{2}\left(\cos(x)\right)^{2}=0$$

y ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$a+ax^{2}y-ax^{2}\left(\cos(x)\right)^{2}=y$$

a'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(1+x^{2}y-x^{2}\left(\cos(x)\right)^{2}\right)a=y$$

Tenglama standart shaklda.

$$\left(-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1\right)a=y$$

Ikki tarafini 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} ga bo‘ling.

$$\frac{\left(-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1\right)a}{-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1}=\frac{y}{-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1}$$

1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} ga bo'lish 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} ga ko'paytirishni bekor qiladi.

$$a=\frac{y}{-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1}$$

y ni 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} ga bo'lish.

$$a=\frac{y}{x^{2}\left(-\left(\cos(x)\right)^{2}+y\right)+1}$$

a qiymati 0 teng bo‘lmaydi.

$$a=\frac{y}{x^{2}\left(-\left(\cos(x)\right)^{2}+y\right)+1}\text{, }a\neq 0$$