
Using the fact that cos is 2π-periodic, we have

That is,
for any
and integer
.

We get 2 solutions in the interval [0, 2π] for
and
,

Answer:
AB-C^2 = 3x^3 + x^2 + 9
Step-by-step explanation:
Hi
AB = (x^2)*(3x+2)= 3x^3 + 2x^2
C^2= (x-3)^2 = x^2 - 9
So
AB-C^2 = 3x^3 + 2x^2 - x^2 + 9 = 3x^3 + x^2 + 9
I only have the answer for 3 and i believe it is 20,090
Answer:
A = 4/7
B = 2/3
C = 2/5
Step-by-step explanation:
It's Easy.