Whakaoti i te a-y/ax^2+y=cos^2x

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

Whakaoti mō a (complex solution)

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

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe a ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te ax^{2}.

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

Tangohia te ax^{2}\left(\cos(x)\right)^{2} mai i ngā taha e rua.

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

Me tāpiri te y ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

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

Pahekotia ngā kīanga tau katoa e whai ana i te a.

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

He hanga arowhānui tō te whārite.

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

Whakawehea ngā taha e rua ki te 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2}.

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

Mā te whakawehe ki te 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} ka wetekia te whakareanga ki te 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2}.

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

Whakawehe y ki te 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2}.

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

Tē taea kia ōrite te tāupe a ki 0.

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

Whakaoti mō a

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

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe a ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te ax^{2}.

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

Tangohia te ax^{2}\left(\cos(x)\right)^{2} mai i ngā taha e rua.

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

Me tāpiri te y ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

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

Pahekotia ngā kīanga tau katoa e whai ana i te a.

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

He hanga arowhānui tō te whārite.

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

Whakawehea ngā taha e rua ki te 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2}.

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

Mā te whakawehe ki te 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} ka wetekia te whakareanga ki te 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2}.

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

Whakawehe y ki te 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2}.

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

Tē taea kia ōrite te tāupe a ki 0.

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