Correct option is A.\(\frac{\pi}3\)

We are given that,

θ = sin^-1{sin (-600°)}

We know that,

sin (2π – θ) = sin (4π – θ) = sin (6π – θ) = sin (8π – θ) = … = -sin θ

As, sin (2π – θ), sin (4π – θ), sin (6π – θ), … all lie in IV Quadrant where sine function is negative.

So,

If we replace θ by 600°, then we can write as

sin (4π – 600°) = -sin 600°

Or, sin (4π – 600°) = sin (-600°)

Or, sin (720° – 600°) = sin (-600°) …(i)

[∵, 4π = 4 × 180° = 720° < 600°]

Thus, we have θ = sin^-1{sin (-600°)}

⇒ θ = sin^-1 {sin (720° – 600°)} [from equation (i)]

⇒ θ = sin^-1 {sin 120°} …(ii)

We know that,

sin (π – θ) = sin (3π – θ) = sin (5π – θ) = … = sin θ

As, sin (π – θ), sin (3π – θ), sin (5π – θ), … all lie in II Quadrant where sine function is positive.

So, If we replace θ by 120°, then we can write as

sin (π – 120°) = sin 120°

Or, sin (180° - 120°) = sin 120° …(iii)

[∵, π = 180° < 120°]

Thus, from equation (ii),

θ = sin^-1{sin 120°}

⇒ θ = sin^-1{sin (180° - 120°)} [from equation (iii)]

⇒ θ = sin^-1 {sin 60°}

Using property of inverse trigonometry,

sin^-1(sin A) = A ⇒ θ = 60°

⇒ θ = 60°

⇒ θ = \(\frac{\pi}3\)