If we assume Hardy-Weinberg equilibrium, we can use the equation p^2 + 2pq + q^2 = 1, where p is the frequency of the dominant allele and q is the frequency of the recessive allele.
Given that the dominant phenotype occurs in 90% of the population, we can say that p^2 + 2pq = 0.9 or 9/10.
We can then solve for p:
p^2 + 2pq = 0.9
p^2 + 2p(1-p) = 0.9
p^2 - 2p^2 + 2p = 0.9
-p^2 + 2p = -0.1
p^2 - 2p + 1 = 1.1
(p-1)^2 = 1.1
p-1 = ±sqrt(1.1)
p = 1 ± sqrt(1.1)
Since p is the frequency of the dominant allele, it cannot be greater than 1. Therefore, we take the negative root:
p = 1 - sqrt(1.1)
p = 0.05
So the frequency of the dominant allele is 0.05 or 5%.