求解 125-5^2y^1/2=5sin(t)

$$125 - 5 ^ { 2 } y ^ { 1 / 2 } = 5 \sin ( t )$$

求解 t 的值 (复数求解)

$$t=-i\ln(i\sqrt{25y-250\sqrt{y}+624}-5i\sqrt{y}+25i)+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}$$

$$t=-i\ln(-i\sqrt{25y-250\sqrt{y}+624}-5i\sqrt{y}+25i)+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}$$

求解 y 的值 (复数求解)

$$\left\{\begin{matrix}y=\frac{\left(-\sin(t)+25\right)^{2}}{25}\text{, }&arg(\frac{-\sin(t)+25}{5})<\pi \\y=0\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }t=2\pi n_{2}-i\ln(4\sqrt{39}i+25i)\text{ or }\exists n_{1}\in \mathrm{Z}\text{ : }t=2\pi n_{1}-i\ln(-4\sqrt{39}i+25i)\end{matrix}\right.$$

求解 y 的值

$$y=\frac{\left(-\sin(t)+25\right)^{2}}{25}$$

$$\frac{-\sin(t)+25}{5}\geq 0$$