Help please?

enter image source here

1 Answer
Mar 6, 2018

#60^@; 120^@; 240^@; 300^@#

Explanation:

  1. #4cos^2 t = 1#
    #cos^2 t = 1/4#
    #cos t = +- 1/2#
    a. #cos t = 1/2#.
    Trig table and unit circle give 2 solutions
    #t = +- 60^@#
    Note. #- 60^@# is co-terminal to #(300^@)#
    b. #cos t = - 1/2#
    Trig table and unit circle -->
    #t = +- 120^@#
    Note. #(- 120^@)# is co-terminal to #(240^@)#
    Answers for (0, 360):
    60, 120, 240 300
  2. #cos^2 t + cos t - 1 = 0#
    #D = d^2 = b^2 - 4ac = 1 + 4 = 5# --># d = +- sqrt5#
    There are 2 real roots:
    #cos t = - b/(2a) +- d/(2a) = 1/2 +- sqrt5/2 = (1 +- sqrt5)/2#
    a. #cos t = (1 + sqrt5)/2# (rejected as > 1
    b. #cos t = -1.236/2 = - 0.62#
    Calculator and unit circle give 2 solutions:
    #t = +- 128^@17#
    Answer for (0, 180):
    #t = 128^@17#