Whakaoti i te tan2x=gx+1

$$\tan 2 x = g x + 1$$

Whakaoti mō g (complex solution)

$$g=-\frac{\left(1+i\right)e^{2ix}+\left(1-i\right)e^{-2ix}}{2x\cos(2x)}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}\text{ and }x\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

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

$$gx+1=\tan(2x)$$

Tangohia te 1 mai i ngā taha e rua.

$$gx=\tan(2x)-1$$

He hanga arowhānui tō te whārite.

$$xg=\tan(2x)-1$$

Whakawehea ngā taha e rua ki te x.

$$\frac{xg}{x}=\frac{\left(-1-i\right)e^{2ix}+\left(-1+i\right)e^{-2ix}}{2\cos(2x)x}$$

Mā te whakawehe ki te x ka wetekia te whakareanga ki te x.

$$g=\frac{\left(-1-i\right)e^{2ix}+\left(-1+i\right)e^{-2ix}}{2\cos(2x)x}$$

Whakawehe \frac{\left(-1-i\right)e^{2ix}+\left(-1+i\right)e^{-2ix}}{2\cos(2x)} ki te x.

$$g=\frac{\left(-1-i\right)e^{2ix}+\left(-1+i\right)e^{-2ix}}{2x\cos(2x)}$$

Whakaoti mō g

$$g=-\frac{\cos(2x)-\sin(2x)}{x\cos(2x)}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}\text{ and }x\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

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

$$gx+1=\tan(2x)$$

Tangohia te 1 mai i ngā taha e rua.

$$gx=\tan(2x)-1$$

He hanga arowhānui tō te whārite.

$$xg=\tan(2x)-1$$

Whakawehea ngā taha e rua ki te x.

$$\frac{xg}{x}=\frac{\sin(2x)-\cos(2x)}{\cos(2x)x}$$

Mā te whakawehe ki te x ka wetekia te whakareanga ki te x.

$$g=\frac{\sin(2x)-\cos(2x)}{\cos(2x)x}$$

Whakawehe \frac{\sin(2x)-\cos(2x)}{\cos(2x)} ki te x.

$$g=\frac{\sin(2x)-\cos(2x)}{x\cos(2x)}$$