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

Algebra/Trig ReviewSince we want the solutions on [0, 2 π ] negative values of n aren’t needed for thisproblem. So, plugging in values of n will give the following four solutions7. cos( 3x ) = 2π 5π 9π 13πt = , , ,8 8 8 8SolutionThis is a trick question that is designed to remind you of certain properties about sineand cosine. Recall that −1 ≤cos( θ ) ≤ 1 and −1 ≤sin ( θ ) ≤ 1. Therefore, since cosinewill never be greater that 1 it definitely can’t be 2. So THERE ARE NOSOLUTIONS to this equation!8. sin ( 2x) = − cos( 2x)SolutionThis problem is a little different from the previous ones. First, we need to do somerearranging and simplification.sin(2 x) = −cos(2 x)sin(2 x)= −1cos(2 x)tan ( 2x)= −1So, solving sin(2 x) = − cos(2 x)is the same as solving tan(2 x ) = − 1. At some levelwe didn’t need to do this for this problem as all we’re looking for is angles in whichsine and cosine have the same value, but opposite signs. However, for otherproblems this won’t be the case and we’ll want to convert to tangent.Looking at our trusty unit circle it appears that the solutions will be,3π2x= + 2 π n, n= 0, ± 1, ± 2, 47π2x= + 2 π n, n= 0, ± 1, ± 2, 4Or, upon dividing by the 2 we get the solutions3πx= + π n, n= 0, ± 1, ± 2, 87πx= + π n, n= 0, ± 1, ± 2, 8No interval was given so we’ll stop here.9. 2sin ( θ) cos( θ ) = 1© 2006 Paul Dawkins 69http://tutorial.math.lamar.edu/terms.aspx