Given the function f (x) = 2Sin (3x + θ), X belongs to [2 α - 5 π, 3 α] and is an odd function, where θ belongs to (0,2 π), what is the value of α - θ? Please write down the specific process of solving the problem,

First, because the function is odd, the domain of definition is symmetric about 0
So, (2 α - 5 π) = 3 α, the solution is α = π
Then f (- x) = - f (x), f (0) = - f (0), that is, f (0) = 0
So, 2Sin θ = 0, that is, sin θ = 0
And theta belongs to (0,2 π), so theta = π
Finally, α - θ = 0

Given the function f (x) = 2Sin (3x + θ), X belongs to [2 α - 5 π, 3 α] and is an odd function, where θ belongs to (0,2 π), what is the value of α - θ? Please write down the specific process of solving the problem,

Given the function f (x) = 2Sin (3x + θ), X belongs to [2 α - 5 π, 3 α] and is an odd function, where θ belongs to (0,2 π), what is the value of α - θ? Please write down the specific process of solving the problem,

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