Whakaoti mō A (complex solution)
\left\{\begin{matrix}A=-\frac{B-y}{\cos(x)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}\\A\in \mathrm{C}\text{, }&y=B\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}\end{matrix}\right.
Whakaoti mō A
\left\{\begin{matrix}A=-\frac{B-y}{\cos(x)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}\\A\in \mathrm{R}\text{, }&y=B\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}\end{matrix}\right.
Whakaoti mō B
B=-A\cos(x)+y
Graph
Tohaina
Kua tāruatia ki te papatopenga
A\cos(x)+B=y
Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.
A\cos(x)=y-B
Tangohia te B mai i ngā taha e rua.
\cos(x)A=y-B
He hanga arowhānui tō te whārite.
\frac{\cos(x)A}{\cos(x)}=\frac{y-B}{\cos(x)}
Whakawehea ngā taha e rua ki te \cos(x).
A=\frac{y-B}{\cos(x)}
Mā te whakawehe ki te \cos(x) ka wetekia te whakareanga ki te \cos(x).
A\cos(x)+B=y
Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.
A\cos(x)=y-B
Tangohia te B mai i ngā taha e rua.
\cos(x)A=y-B
He hanga arowhānui tō te whārite.
\frac{\cos(x)A}{\cos(x)}=\frac{y-B}{\cos(x)}
Whakawehea ngā taha e rua ki te \cos(x).
A=\frac{y-B}{\cos(x)}
Mā te whakawehe ki te \cos(x) ka wetekia te whakareanga ki te \cos(x).
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