Help please?
1 Answer
Mar 6, 2018
Explanation:
#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#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#