the distance in kilometres of a streetcar away from the station is modellled by d(t)= - 6 sin (pi/10(t+5))+6 where t is the time in minutes. each round trip take 20 minutes. When will the streetcar be 8km away from the station in the next 30 mins
"- 6 sin (\u03c0\/10(t+5))+6=8 \\\\ - 6 sin (\u03c0\/10(t+5))=8-6 \\\\ - 6 sin (\u03c0\/10(t+5))=2 \\\\ sin (\u03c0\/10(t+5))=-2\/6 \\\\ sin (\u03c0\/10(t+5))=-1\/3 \\\\\n\u03c0\/10(t+5))=(-1)^k arcsin(-1\/3) +\u03c0*k, \\ k\u2208Z \\\\ t+5=10*((-1)^k arcsin(-1\/3) +\u03c0*k)\/\u03c0, \\ k\u2208Z \\\\ t=10*((-1)^k arcsin(-1\/3) +\u03c0*k)\/\u03c0 -5, \\ k\u2208Z \\\\ \nt=10*((-1)^{k+1} arcsin(1\/3) +\u03c0*k)\/\u03c0 -5, \\ k\u2208Z \\\\ \nt=\\mp10arcsin(1\/3)\/\u03c0 -5+20k, \\ k\u2208Z \\\\ \nt\\approx\\mp6.08173+20k, \\ k\u2208Z \\\\ k=0 ,\\ t\\approx6.1 \\ (min) \\\\\n k=1 ,\\ t\\approx 13.9 \\ (min) \\\\\n k=1 ,\\ t\\approx26.1\\ (min) \\\\ Answer: \\ 6.1 min,\\ 13.9 min, \\ 26.1 min"
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