Question 769962: 1) Find the smallest positive value of t for which f(t) = 2 sin (2t − π/6) is the following.
f(t) attains a maximum value.
t =
2) Find all values of t in the interval [0, 2π] satisfying the given equation. (Enter your answers as a comma-separated list.)
sqrt(6) sin 2t + 3 tan 2t = 0
t =
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 1) Find the smallest positive value of t for which f(t) = 2 sin (2t − π/6) is the following.
f(t) attains a maximum value.
sin is maximum at π/2
sin(2t-π/6)=sin(π/2)
2t-π/6=π/2
2t=π/2+π/6=4π/6=2π/3
t = π/3
f(t) attains a maximum value when t=π/3
..
2) Find all values of t in the interval [0, 2π] satisfying the given equation. (Enter your answers as a comma-separated list.)
sqrt(6) sin 2t + 3 tan 2t = 0
√6sin(2t)+(3sin(2t)/cos(2t))=0
LCD:cos(2t)
√6sin(2t)cos(2t)+3sin(2t)=0
sin(2t)(√6cos(2t)+3)=0
..
sin(2t)=0
2t=0, π,2π
t=0, π/2,π
..
√6cos(2t)+3)=0
√6cos(2t)=-3
cos(2t)=-3/√6≈-1.22... (reject, -1 ≤ cos(2t) ≤ 1)
..
values of t in the interval [0, 2π]=0, π, 2π
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