Answer:
A sample of 6758 would be needed.
Explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
The margin of error is:

The newspaper took a random sample (assume it is an SRS) of 1200 registered voters and found that 580 would vote to raise taxes.
This means that

90% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
How large a sample n would you need to estimate p with margin of error 0.01 with 90% confidence?
This is n for which M = 0.01. So






Rounding up:
A sample of 6758 would be needed.