Question #3ec6c

1 Answer
Feb 27, 2018

#218^@32 + k360^@#
#321^@68 + k360^@#

Explanation:

#sin t = cos^2 t#
#sin t = 1 - sin^2 t#
#sin^2 t + sin t - 1 = 0#
Solve this quadratic equation for sin t.
#D = d^2 = b^2 - 4ac = 1 + 4 = 5# --> #d = +- sqrt5#
There are 2 real roots:
#sin t = -b/(2a) +- d/(2a) = - 1/2 +- sqrt5/2 = (1 +- sqrt5)/2#
a. #sin t = (1 + sqrt5)/2 = 3.24/2 = 1.62# (rejected as > 1)
b. #sin t = (1 - sqrt5)/2 = -1.24/2 = - 0.62#
Calculator and unit circle give 2 solutions:
#t = 218^@32#, and #t = - 38^@32#, or #t = 321^@68# (co-terminal)
For general answers, add #k360^@#