Question

1)
Find function domain
f(x) = sqrt( 2sin x - 1 )

Answer 1

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

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}