Algebra/Trig Review - Pauls Online Math Notes - Lamar University

Algebra/Trig Review4π2πn− 4x= + , n= 0, ± 1, ± 2, 15 55π2πn− 4x= + , n= 0, ± 1, ± 2, 15 5It’s somewhat messy, but it is the solution. Don’t get excited when solutions getmessy. They will on occasion and you need to get used to seeing them.5. 2sin ( 3x ) = 1 on [ − ππ , ]SolutionI’m going to leave most of the explanation that was in the previous three out of thisone to see if you have caught on how to do these.( x)2sin 3 = 11sin ( 3x)=2By examining a unit circle we see thatπ3x= + 2 π n, n= 0, ± 1, ± 2, 65π3x= + 2 π n, n= 0, ± 1, ± 2, 6Or, upon dividing by 3,π 2πnx= + , n= 0, ± 1, ± 2, 18 35π2πnx= + , n= 0, ± 1, ± 2, 18 3− ππ , . So, let’s start trying someNow, we are looking for solutions in the range [ ]values of n.n = 0 :n = 1 :n = 2 :π 2π 13πx = + = < π so a solution18 3 185π 2π 17πx = + = < π so a solution18 3 18π 4π 25πx = + = > π so NOT a solution18 3 185π 4π 29πx = + = > π so NOT a solution18 3 18π5πx= & x=18 18© 2006 Paul Dawkins 67http://tutorial.math.lamar.edu/terms.aspx