Find three cube roots for the following complex number. Leave your answers in trigonometric form. (Enter your answers from smallest to largest angle measure.)8(cos 303° + i sin 303°)

To find the cube roots of a complex number in trigonometric form, we use De Moivre's Theorem. The formula for the nth roots of a complex number in trigonometric form is:

r^(1/n)[cos((θ+2πk)/n) + i sin((θ+2πk)/n)]

where r is the modulus of the complex number, θ is the argument of the complex number, n is the root you want to find (in this case, 3 for cube root), and k is an integer from 0 to n-1.

Given the complex number 8(cos 303° + i sin 303°), we have r = 8 and θ = 303°.

Step 1: Find r^(1/n) The cube root of 8 is 2.

Step 2: Substitute r, θ, n, and k into the formula and simplify to find the cube roots.

For k = 0: 2[cos((303°+2π(0))/3) + i sin((303°+2π(0))/3)] = 2[cos(101°) + i sin(101°)]

For k = 1: 2[cos((303°+2π(1))/3) + i sin((303°+2π(1))/3)] = 2[cos(221°) + i sin(221°)]

For k = 2: 2[cos((303°+2π(2))/3) + i sin((303°+2π(2))/3)] = 2[cos(341°) + i sin(341°)]

So, the three cube roots of 8(cos 303° + i sin 303°) are 2[cos(101°) + i sin(101°)], 2[cos(221°) + i sin(221°)], and 2[cos(341°) + i sin(341°)].