( 3 \alpha - 4 ) = - \frac { 1 } { 2 } \text { for } 0 \leq \alpha \leq 2 \tau
Solve for τ
\left\{\begin{matrix}\tau \geq \frac{\alpha }{2}\text{, }&\alpha \geq 0\\\tau \in \mathrm{R}\text{, }&\alpha =-\frac{f}{6}+\frac{4}{3}\end{matrix}\right.
Solve for α
\left\{\begin{matrix}\\\alpha =-\frac{f}{6}+\frac{4}{3}\text{, }&\text{unconditionally}\\\alpha \in \begin{bmatrix}0,2\tau \end{bmatrix}\text{, }&\tau \geq 0\end{matrix}\right.
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