Find $t$ in $i = 50\sin\left(120\pi t -\frac{3\pi}{25}\right)$ where $i = 25$ 
An alternating current generator produces a current given by the equation
  $$i = 50\sin\left(120\pi t - \frac{3\pi}{25}\right)$$
  where $t$ is the $\text{time}$ in $\text{seconds}$.
(Q) Find the first two values of $t$ when $i = 25 \text{ amperes}$
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Is my working correct? Kinda stuck with this problem, what do I do next?
$50\sin(120\pi t-3\pi/25)= 25$
$\sin(120\pi t-3\pi/25)= 25/50$
$\sin(120\pi t-3\pi/25)= 0.5$
$\text{Ref angle }= \sin^{-1}(0.5)$
Since $\sin(120\pi t-3\pi/25)= 0.5 > 0$, it is at $1^{\text{st}}$ or $2^{\text{nd}}$ quadrant
$(120\pi t-3\pi /25)=\pi/6$
 A: You're right so far =) 
Now, you need to consider the angle which lies in Quadrant II to obtain your $2^{\text{nd}}$ value for $t$
You could use the fact $\sin\theta = \sin(\pi - \theta)$ because if $\theta \in \left(0,\frac{\pi}{2}\right)$, then $\pi - \theta \in \left(\frac{\pi}{2} , \pi\right)$
Just to stimulate analytical thinking of trigonometric and inverse trigonometric functions, let us consider a general equation  for alternating current:
$$i = \mathcal A\sin\left(\omega t + \phi\right) \implies t = \frac{1}{\omega}\left( \arcsin \frac{i}{ \mathcal A} - \phi\right)$$
Here, in the calculation of $t$, we have constricted $\frac{i}{\mathcal A} \in \left[-1, 1\right]$ inorder to use the $\arcsin(x)$ function which only spits out angles in the first quadrant        ( from $0 \text{ rad}$ to $\frac{\pi}{2} \text{ rad}$) 
This gives us one value for $t$ but what about the second answer? 
Well, consider $\omega t + \phi \in \left(\frac{\pi}{2}, \pi\right)$ . For which value in that range would $\sin(\omega t + \phi) = \sin(\arcsin \frac{i}{\mathcal A})$ ?
Use the fact that $\sin\theta = \sin\alpha \iff \theta = n\pi + (-1)^n\alpha\  , n \in \mathbb  Z$ in such a way as to get your required angle which lies in the second quadrant.
But note this, we could have effortlessly developed a general equation for all $t$ by considering that:
$$\sin(y) = x \iff y = n\pi + (-1)^n\arcsin x$$
Hence,
$$t_n = \frac{1}{\omega}\left(n\pi + (-1)^n\arcsin \frac{i}{\mathcal A} - \phi\right)\ ,\, n\in\Bbb Z$$
Make the substitutions
$
\begin{align}
\omega &= 120\pi \text{ rad/s}\\
i &= 25 \text{ amp}\\
\mathcal A &= 50 \text{ amp}\\
\phi &= - \frac{3\pi}{25} \text{ rad}
\end{align}
$
and find the values of $t_{(n=1)}$ and $t_{(n=2)}$
A: Hint: Let $x=120\pi t-3\pi/25$. If $x=\theta $ is one
solution of the equation $y=\sin x$ then another solution is $x=\pi -\theta $.(*)
Complete solution using your work. If the hint isn't enough, hover your mouse over the grey to see
You have correctly found the first angle $\theta >0$ for which $\sin \theta =1/2$. It's $\arcsin 1/2=\pi /6\in \left[0,\pi /2\right] $. 

 The first value of $t$ when $i=25\,  \textrm{A}$ (I denote $t_{1}$) satisfies your last equation.

\begin{equation*}
120\pi t-\frac{3\pi }{25}=\frac{\pi }{6}.
\end{equation*}

 Simplify by dividing both sides by $\pi $\begin{equation*}120t-\frac{3}{25}=\frac{1}{6}\end{equation*}and solve it

\begin{equation*}
t=t_{1}=\frac{1/6+3/25}{120}=\frac{43}{18\,000}\text{ }\mathrm{s}\approx
2.389\text{ }\mathrm{ms}
\end{equation*}
Since in general $\sin x=\sin \left( \pi -x\right) $, we have 

 \begin{equation*}\sin (\pi /6)=\sin \left( \pi -\pi /6\right) =\sin (5\pi /6).\end{equation*} 

Hence $\theta _{2}=5\pi /6\in \left[ \pi /2,\pi \right]$ is the second angle $\theta>0$ for which $\sin \theta =1/2$. 

 As a consequence, and similarly to $t_{1}$, the second value of $t$ when $i=251 \, \textrm{A}$ (I denote $t_{2}$), is the solution of the equation 

\begin{equation*} 120\pi t-\frac{3\pi }{25}=\frac{5\pi }{6},\end{equation*}

 which is 

\begin{equation*}
t=t_{2}=\frac{5/6+3/25}{120}\text{ }\mathrm{s}=\frac{143}{18\,000}\approx
7.944\text{ }\mathrm{ms}
\end{equation*}
--
(*) The general form for the solutions $x$ of $y=\sin x$ combine both
solutions $x=\theta ,x=\pi -\theta $ added to any multiple of $2\pi $. So $
x=2k\pi +\theta ,k\in \mathbb{Z}
$ and $x=\pi -\theta +2k\pi =(2k+1)\pi -\theta ,k\in \mathbb{Z}$ are solutions, which can be written in a single expression as $x=n\pi
+(-1)^{n}\theta $, $\theta \in \left[ -\frac{1}{2},\frac{1}{2}\right] $ and $n\in 
\mathbb{Z}.$
