Find the least positive value of t for which d = 0.

Question

d=4cos8𝜋/5(t)

0=4cos8𝜋/5(t)

Answer 1

cosines are zero when the angle = pi/2

cos(8pi/5t) =cos(pi/2)

8pi/5t = pi/2 the pi's cancel

8/5t = 1/2 cross multiply to get

5t =16 divide by 5

t =16/5

Answer 2

4cos(8πt/5) = 0,cosπ(8πt/5) = 0; 8πt/5 = π/2 + πn, t = 5/16 + 5n/8 and minimum psitive value for t we get when n=0, so, t_min = 5/16