Finding times in first cycle with given voltage
You have:
$$\begin{align*}
V_t &amp;= 100\: \textrm{V}\cdot\operatorname{sin}\left(2 \pi\:\operatorname{rad}\cdot25\operatorname{Hz}\cdot\: t - 0.523 \operatorname{rad}\right)
\end{align*}$$

And you want to solve for \$t\$ where \$t\ge 0\$ and \$V_t=-40\:\textrm{V}\$. So, let's set \$x\$ as follows:

$$ x= 2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}\cdot\: t - 0.523 \operatorname{rad}$$

Then we have:

$$\begin{align*}
-40\:\textrm{V} &amp;= 100\: \textrm{V}\cdot\operatorname{sin}\left(x\right)\\\\
\frac{-40\:\textrm{V}}{100\: \textrm{V}} &amp;= \operatorname{sin}\left(x\right)\\\\
-0.4 &amp;= \operatorname{sin}\left(x\right)
\end{align*}$$

Before going any further, the value of \$x\$ has multiple solutions. The solutions are:

$$\begin{align*}
x&amp;=2\pi\cdot n + \operatorname{sin}^{-1}\left(-0.4\right)= 2\pi\cdot n - 0.411516846 \operatorname{rad}\\&amp;=2\pi\cdot n +\pi- \operatorname{sin}^{-1}\left(-0.4\right)= 2\pi\cdot n +3.5531095 \operatorname{rad}
\end{align*}$$

Combining this information, we have:

$$\begin{align*}
2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}\cdot\: t - 0.523 \operatorname{rad}&amp;= 2\pi\cdot n - 0.411516846 \operatorname{rad}\\2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}\cdot\: t - 0.523 \operatorname{rad}&amp;= 2\pi\cdot n +3.5531095\operatorname{rad}
\end{align*}$$

These solve out as:



$$\begin{align*}
t&amp;= \frac{2\pi\cdot n - 0.411516846 \operatorname{rad} + 0.523 \operatorname{rad}}{2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}}=\frac{2\pi\cdot n +0.111483154\operatorname{rad}}{2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}}\\\\
t&amp;= \frac{2\pi\cdot n +3.5531095 \operatorname{rad} + 0.523 \operatorname{rad}}{2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}}=\frac{2\pi\cdot n +4.0761095\operatorname{rad}}{2 \pi\:\operatorname{rad}\cdot\:25\operatorname{Hz}}
\end{align*}$$



That's the full answer. But for values of \$t\ge 0\$, you find the following for the first two answers with \$n=0\$:

$$\begin{align*}
t&amp;=709.723801\:\mu\textrm{s}\\\\
t&amp;=25.949319\:\textrm{ms}
\end{align*}$$

I hope that helps out. As you can see, the trick is mostly in taking very careful steps and not to move too rapidly towards a quick "calculator" solution, which would find the first answer, perhaps, but not the second (which appears to be the desired one.)