Question

определите область значения функции y=3+2sin(x-п/6)
варианты:
[2;3]
[-3;2]
[-1;1]
[1;5]

Answer 1

$-1 leq sin alpha leq 1; ; to ; ; -1 leq sin(x-frac{pi}{6}) leq 1\\-2 leq 2sin(x-frac{pi}{6}) leq 2\\3-2 leq 3+2sin(x-frac{pi}{6} leq 2+3\\1 leq 3+2sin(x-frac{pi}{6}) leq 5\\ yin [1,5]$