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10 April, 00:29

1)

Find function domain

f (x) = sqrt (2sin x - 1)

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Answers (1)
  1. 10 April, 03:02
    0
    {x ∈ ℝ : x ≥ π/6 + 2πn and x ≤ π/6 + 2πn and n ∈ ℤ}

    Step-by-step explanation:

    sinx can run from - 1 to + 1

    2sinx can run from - 2 to + 2

    2sinx - 1 can run from - 3 to + 1

    However, the square root is imaginary when x < 0. So, the condition is

    2sinx - 1 ≥ 0

    2sinx ≥ 1

    sinx ≥ ½

    x ≥ π/6 (30°)

    So, in the interval [0, 2π], π/6 ≤ x ≤ 5π/6

    However, the sine is a cyclic function and repeats itself every 2π.

    Over all real numbers, the condition is (π/6 + 2πn) ≤ x ≤ (5π/6 + 2πn).

    The domain is then

    {x ∈ ℝ : x ≥ π/6 + 2πn and x ≤ π/6 + 2πn and n ∈ ℤ}
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