SOLUTION: For what values of t on the interval [0, 2pi] is sin t = {{{sqrt( 3 )/2}}}

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Question 753622: For what values of t on the interval [0, 2pi] is sin t = sqrt%28+3+%29%2F2
Answer by lwsshak3(11628) About Me  (Show Source):
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For what values of t on the interval [0, 2pi] is sin t = sqrt%28+3+%29%2F2
sin(t)=√3/2
t=π/3, 2π/3