\left\{ \begin{array} { l } { \sqrt[ 3 ] { x ^ { 2 } } \leq y + 1 } \\ { \sin \frac { y } { 2 } = x ^ { 3 } \cdot \tan y } \end{array} \right.
Solve for x
\left\{\begin{matrix}x\in \begin{bmatrix}-\left(y+1\right)^{\frac{3}{2}},\left(y+1\right)^{\frac{3}{2}}\end{bmatrix}\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }y=2\pi n_{1}\text{, }n_{1}>-1\\x=\frac{2^{\frac{2}{3}}\sqrt[3]{\frac{\cos(y)}{\cos(\frac{y}{2})}}}{2}\text{, }&\nexists n_{4}\in \mathrm{Z}\text{ : }y=\frac{\pi \left(2n_{4}+1\right)}{2}\text{ and }\nexists n_{3}\in \mathrm{Z}\text{ : }y=\pi \left(2n_{3}+1\right)\text{ and }y\geq -1\text{ and }\exists n_{2}\in \mathrm{Z}\text{ : }\left(y>\frac{\pi n_{2}}{2}\text{ and }y<\frac{\pi \left(n_{2}+1\right)}{2}\right)\text{ and }\frac{\sqrt[3]{|\cos(y)|}}{\sqrt[3]{2|\cos(\frac{y}{2})|}}\leq \left(y+1\right)^{\frac{3}{2}}\end{matrix}\right.
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