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Damari Vaughn
Mathematics
11 June, 08:44
Find function domain
f (x) = sqrt (2sin x - 1)
+1
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Whispy
11 June, 09:03
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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