Answer:
Step-by-step explanation:
4 and 6
<span>we have that
the cube roots of 27(cos 330° + i sin 330°) will be
</span>∛[27(cos 330° + i sin 330°)]
we know that
e<span>^(ix)=cos x + isinx
therefore
</span>∛[27(cos 330° + i sin 330°)]------> ∛[27(e^(i330°))]-----> 3∛[(e^(i110°)³)]
3∛[(e^(i110°)³)]--------> 3e^(i110°)-------------> 3[cos 110° + i sin 110°]
z1=3[cos 110° + i sin 110°]
cube root in complex number, divide angle by 3
360nº/3 = 120nº --> add 120º for z2 angle, again for z3
<span>therefore
</span>
z2=3[cos ((110°+120°) + i sin (110°+120°)]------ > 3[cos 230° + i sin 230°]
z3=3[cos (230°+120°) + i sin (230°+120°)]--------> 3[cos 350° + i sin 350°]
<span>
the answer is
</span>z1=3[cos 110° + i sin 110°]<span>
</span>z2=3[cos 230° + i sin 230°]
z3=3[cos 350° + i sin 350°]<span>
</span>
Answer:
x≥4
Step-by-step explanation:
We want to solve for x in
Let us add 3 to both sides:
We simplify to get:
Divide through by 2 to get:
