In a population frequency of the heterozygous individual is 42% the find out an individual with recessive phenotype and homozygous dominant phenotype.
In a population, the frequency of heterozygous individual is 42%. It means 2pq = 0.42. Hence, pq = 0.21 = 0.7 x 0.3. Hence, we can say that p = 0.7 and q = 0.3.  Now, the frequency of homozygous dominant phenotype = p$$^2$$ = 0.7$$^2$$ = 0.49 = 49%. Similarly, the frequency of homozygous recessive phenotype = q$$^2$$ = 0.3$$^2$$ = 0.9 = 9%. Hence, frequency of individual with recessive phenotype is 9% and frequency of individual with homozygous dominant phenotype is 49%.