E(t)E, left parenthesis, t, right parenthesis models the daily amount of energy (in kilojoules, \text{kJ}kJstart text, k, J, end text) that a solar panel in Pago Pago generates, ttt days after the autumn equinox. Here, ttt is entered in radians. E(t) = {624}\sin\left({\dfrac{2\pi}{365}}t \right) + {8736}E(t)=624sin( 365 2π t)+8736E, left parenthesis, t, right parenthesis, equals, 624, sine, left parenthesis, start fraction, 2, pi, divided by, 365, end fraction, t, right parenthesis, plus, 8736 What is the first day after the autumn equinox that the solar panel generates 8400\text{ kJ}8400 kJ8400, start text, space, k, J, end text?

216 days

Step-by-step explanation:

Converting the problem into mathematical terms

E(t) = {624}\sin\left({\maroonD{\dfrac{2\pi}{365}}}t\right) + {8736}E(t)=624sin(  

365

2π

t)+8736E, left parenthesis, t, right parenthesis, equals, 624, sine, left parenthesis, start color #ca337c, start fraction, 2, pi, divided by, 365, end fraction, end color #ca337c, t, right parenthesis, plus, 8736 has a period of \dfrac{2\pi}{\maroonD{\scriptsize\dfrac{2\pi}{365}}}=365  

365

2π

2π

=365start fraction, 2, pi, divided by, start color #ca337c, start fraction, 2, pi, divided by, 365, end fraction, end color #ca337c, end fraction, equals, 365 days.

We want to find the first solution to the equation E(t)=8400E(t)=8400E, left parenthesis, t, right parenthesis, equals, 8400 within the period 0<t<3650<t<3650, is less than, t, is less than, 365.