Originally Posted by
UWstudent The voltage output (in volts) of an electrical circuit at time t seconds is given by the following function.
V(t) = 2^{2 sin(5πt − 3π) + 1 }(e) During the first second, determine when the voltage output of the circuit is 3 volts. (Enter your answers as a comma-separated list. Round your answers to four decimal places.)
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My approach is to
1. set the function equal to 3 }3=2^{2}^{sin(5∏t-3∏)+1
2. take the natural log to get rid of the exponent }ln3=(2sin(5∏t-3∏)+1)ln2
[SUP]
(ln3/ln2)-1=2sin(5∏t-3∏)
[(ln3/ln2)-1]/2=sin(5∏t-3∏)
arcsin ([(ln3/ln2)-1]/2)=5∏t-3∏
t= (arcsin([ln3/ln2)-1]/2) + 3∏) / 5∏
t=0.6189
I don't know if I'm on the right track so far, and I have no idea where to go from here? Any help would be soooooo great!! Thanks.