解 y (復數求解)
y=\arctan(\frac{-\sqrt{2\left(\cos(2x)+1\right)}+2}{2\sin(x)})
\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}
解 x
x=arcSin(2SinI(y)CosI(y))+2\pi n_{201}\text{, }n_{201}\in \mathrm{Z}\text{, }\exists n_{32}\in \mathrm{Z}\text{ : }\left(n_{32}<\left(arcSin(2SinI(y)CosI(y))+2\pi n_{201}\right)\pi ^{-1}\text{ and }n_{32}>\left(arcSin(2SinI(y)CosI(y))+2\pi n_{201}+\left(-1\right)\pi \right)\pi ^{-1}\right)
x=\pi +2\pi n_{202}+\left(-1\right)arcSin(2SinI(y)CosI(y))\text{, }n_{202}\in \mathrm{Z}\text{, }\exists n_{32}\in \mathrm{Z}\text{ : }\left(n_{32}<\left(\pi +2\pi n_{202}+\left(-1\right)arcSin(2SinI(y)CosI(y))\right)\pi ^{-1}\text{ and }n_{32}>\left(2\pi n_{202}+\left(-1\right)arcSin(2SinI(y)CosI(y))\right)\pi ^{-1}\right)
解 y
y=\arctan(\frac{-\sqrt{-\left(\sin(x)\right)^{2}+1}+1}{\sin(x)})
\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}
圖表
共享
已復制到剪貼板
示例
二次方程式
{ x } ^ { 2 } - 4 x - 5 = 0
三角學
4 \sin \theta \cos \theta = 2 \sin \theta
線性方程
y = 3x + 4
算術
699 * 533
矩陣
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
聯立方程
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
微分
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
積分
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
限制
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}