Whakaoti i te y=txsin(2t-1)

$$y = t x \sin ( 2 t - 1 )$$

Whakaoti mō x (complex solution)

$$\left\{\begin{matrix}x=-\frac{2ye^{i-2it}}{t\left(i-ie^{2\left(-2it+i\right)}\right)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }t=\frac{\pi n_{1}}{2}+\frac{1}{2}\text{ and }t\neq 0\\x\in \mathrm{C}\text{, }&\left(\exists n_{1}\in \mathrm{Z}\text{ : }t=\frac{\pi n_{1}}{2}+\frac{1}{2}\text{ or }t=0\right)\text{ and }y=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$tx\sin(2t-1)=y$$

He hanga arowhānui tō te whārite.

$$t\sin(2t-1)x=y$$

Whakawehea ngā taha e rua ki te t\sin(2t-1).

$$\frac{t\sin(2t-1)x}{t\sin(2t-1)}=\frac{y}{t\sin(2t-1)}$$

Mā te whakawehe ki te t\sin(2t-1) ka wetekia te whakareanga ki te t\sin(2t-1).

$$x=\frac{y}{t\sin(2t-1)}$$

Whakaoti mō x

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

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$tx\sin(2t-1)=y$$

He hanga arowhānui tō te whārite.

$$t\sin(2t-1)x=y$$

Whakawehea ngā taha e rua ki te t\sin(2t-1).

$$\frac{t\sin(2t-1)x}{t\sin(2t-1)}=\frac{y}{t\sin(2t-1)}$$

Mā te whakawehe ki te t\sin(2t-1) ka wetekia te whakareanga ki te t\sin(2t-1).

$$x=\frac{y}{t\sin(2t-1)}$$