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**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.