Réitigh do 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.
Réitigh do 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.
Tráth na gCeist
Trigonometry
5 fadhbanna cosúil le:
\frac { a - y } { a x ^ { 2 } } + y = \cos ^ { 2 } x
Roinn
Cóipeáladh go dtí an ghearrthaisce
a-y+ax^{2}y=ax^{2}\left(\cos(x)\right)^{2}
Ní féidir leis an athróg a a bheith comhionann le 0 toisc nach bhfuil an roinnt faoi nialas sainithe. Méadaigh an dá thaobh den chothromóid faoi ax^{2}.
a-y+ax^{2}y-ax^{2}\left(\cos(x)\right)^{2}=0
Bain ax^{2}\left(\cos(x)\right)^{2} ón dá thaobh.
a+ax^{2}y-ax^{2}\left(\cos(x)\right)^{2}=y
Cuir y leis an dá thaobh. Is ionann rud ar bith móide nialas agus a shuim féin.
\left(1+x^{2}y-x^{2}\left(\cos(x)\right)^{2}\right)a=y
Comhcheangail na téarmaí ar fad ina bhfuil a.
\left(-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1\right)a=y
Tá an chothromóid i bhfoirm chaighdeánach.
\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}
Roinn an dá thaobh faoi 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2}.
a=\frac{y}{-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1}
Má roinntear é faoi 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} cuirtear an iolrúchán faoi 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} ar ceal.
a=\frac{y}{x^{2}\left(-\left(\cos(x)\right)^{2}+y\right)+1}
Roinn y faoi 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}\text{, }a\neq 0
Ní féidir leis an athróg a a bheith comhionann le 0.
a-y+ax^{2}y=ax^{2}\left(\cos(x)\right)^{2}
Ní féidir leis an athróg a a bheith comhionann le 0 toisc nach bhfuil an roinnt faoi nialas sainithe. Méadaigh an dá thaobh den chothromóid faoi ax^{2}.
a-y+ax^{2}y-ax^{2}\left(\cos(x)\right)^{2}=0
Bain ax^{2}\left(\cos(x)\right)^{2} ón dá thaobh.
a+ax^{2}y-ax^{2}\left(\cos(x)\right)^{2}=y
Cuir y leis an dá thaobh. Is ionann rud ar bith móide nialas agus a shuim féin.
\left(1+x^{2}y-x^{2}\left(\cos(x)\right)^{2}\right)a=y
Comhcheangail na téarmaí ar fad ina bhfuil a.
\left(-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1\right)a=y
Tá an chothromóid i bhfoirm chaighdeánach.
\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}
Roinn an dá thaobh faoi 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2}.
a=\frac{y}{-x^{2}\left(\cos(x)\right)^{2}+yx^{2}+1}
Má roinntear é faoi 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} cuirtear an iolrúchán faoi 1+x^{2}y-x^{2}\left(\cos(x)\right)^{2} ar ceal.
a=\frac{y}{x^{2}\left(-\left(\cos(x)\right)^{2}+y\right)+1}
Roinn y faoi 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}\text{, }a\neq 0
Ní féidir leis an athróg a a bheith comhionann le 0.
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