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

Answer:{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}