1)

Find function domain

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

{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}