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

Algebra/Trig ReviewThis problem is almost identical to the previous except now I want all the solutionsthat fall in the interval [ − 2 π,2 π]. So we will start out with the list of all possiblesolutions from the previous problem.π+ 2 π n, n= 0, ± 1, ± 2, ± 3, 611π+ 2 π n, n= 0, ± 1, ± 2, ± 3, 6Then start picking values of n until we get all possible solutions in the interval.First notice that since both the angles are positive adding on any multiples of 2π (npositive) will get us bigger than 2π and hence out of the interval. So, all positivevalues of n are immediately out. Let’s take a look at the negatives values of n.n = −1π11π+ 2π( − 1)=− >−2π6 611ππ+ 2π( − 1)=− >−2π6 6These are both greater than −2πand so are solutions, but if we subtract another 2πoff (i.e use n = − 2) we will once again be outside of the interval.So, the solutions are :π 11 11, π , − π , − π .6 6 6 63. 2sin ( 5x ) = − 3SolutionThis one is very similar to Problem 1, although there is a very important difference.We’ll start this problem in exactly the same way as we did in Problem 1.2sin(5 x) = − 3− 3sin(5 x)=23So, we are looking for angles that will give − out of the sine function. Let’s2again go to our trusty unit circle.© 2006 Paul Dawkins 64http://tutorial.math.lamar.edu/terms.aspx