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

Algebra/Trig ReviewI didn’t put in the x, or cosine value, in the unit circle since it’s not needed for theproblem. I did however note that they will be the same value, except for the negativesign. The angle in the second quadrant will then be,π − 0.2013579 = 2.9402348So, let’s put all this together.2x= 0.2013579 + 2 π n, n= 0, ± 1, ± 2, 2x= 2.9402348 + 2 π n, n= 0, ± 1, ± 2, Note that I added the 2π n onto our angles as well since we know that will be neededin order to get all the solutions. The final step is to then divide both sides by the 2 inorder to get all possible solutions. Doing this gives,x= 0.10067895 + π n, n= 0, ± 1, ± 2, x= 1.4701174 + π n, n= 0, ± 1, ± 2, The answers won’t be as “nice” as the answers in the previous problems but therethey are. Note as well that if we’d been given an interval we could plug in values of nto determine the solutions that actually fall in the interval that we’re interested in.⎛ x ⎞13. 4cos⎜⎟ = − 3⎝5⎠© 2006 Paul Dawkins 73http://tutorial.math.lamar.edu/terms.aspx