Calculus
posted by Mishaka .
Find the three zeros for the following function on the interval 5 </= x </= 5
(1 + 50sin(x)) / (x^2 + 3)

For the expression to be zero,
1 + 50sinx = 0
sinx = 1/50
x = π +.02 or 2π  .02
x = 3.1616 or 6.263
the period of 50sinx is 2π, so adding or subtracting 2π to the above answers will produce more answers
3.1616+2π > beyond domain
3.16162π = 3.1216
6.2632π = .02
so for 5 ≤ x ≤ 5
x = 3.1216 , .02, 3.1216 
Thank you, this helps tremendously!

How did you come up with 3.1216 positive. I've tried to come up with this number but I can't figure it out. Also, it doesn't make 1 + 50 sin(x) = 0.

It was a copy error.
As you can see from the 5th line, I had x = 3.1616
so the final answers are
x = 3.1216 , .02, and 3.1616
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