How do you solve #sec (1/2 alpha) = 4#?

Question

Assume #0° < alpha < 180°#.

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Answer 1 · 2017-02-28T14:27:12

#18^@95 + k720^@#
#310^@04 + k720^@#

#sec (x/2) = 1/(sin (x/2)) = 4#
#sin (x/2) = 1/4#
a. Calculator gives:
#sin (x/2) = 1/4# --> #x/2 = 14^@48 + k360^@#-->
#x = 28^@95 + k720^@#

b. Unit circle gives another (x/2) that has the same sine value (1/4):
#x/2 = 180 - 14.48 = 165^@52 + k360^@#
#x = 310^@04 + k720^@#