Finding first time when a voltage is -30

Question

I have the question "The voltage in an alternating current circuit at any time $t$ seconds is given by $v = 60\sin(40t)$ Volts. Find the first time when the voltage is $-30 V$."

So I set the equation to $-30 = 60\sin(40t)$.

Then $-30/60 = \sin(40t)$.

$40t = \sin^{-1} ( -30/60)$

Therefore $t = -30/40 = -0.75$ seconds.

However the solutions say that $t$ should be $91.63$ ms.

So what have I done wrong ?

Answer 1

Note first that

$$\sin(40t)=-\frac{30}{60}=-\frac12 \tag 1$$

We don't want to blindly take the arcsine of both sides of $(1)$ since the result will be $t<0$.

Rather, we want

$$40t =\pi +\arcsin(1/2)\implies t=\frac{\pi +\pi/6}{40}=\frac{7\pi}{240}\approx 91.63\,\,\text{ms}$$

Answer 2

Hint arcsin should give negative value as time cant be negative!. Now the ac would first give a positive voltage till half time period. Now $\omega=40$ thus $time period =\frac {2\pi}{40} $.half of it is positive voltage. So that time is so $\frac{\pi}{40} $ now time for voltage to go $-30$ from $0$ is $\frac {\pi}{6\times 40} $. This should be added with initial half time period thus using calculator net time comes out approximately as $91.63 ms $