How do I solve sec(2x)-7=0 for the four smallest positive solutions?

More specifically, I have no clue how to get rid of the #sec# of the equation.

1 Answer
Jun 27, 2018

#40^@89; 139^@11#

Explanation:

sec 2x = 7
cos 2x = 1/(sec 2x) = 1/7
Calculator and unit circle give 2 solutions for 2x:
#2x = +- 81^@78 + k360^@#
a. #2x = 81^@78 + k360^@#
#x = 40^@89 + k180#
If k = 0 --> #x = 40^@89#,
If k = 1 --># x = 40.89 + 180 = 220^@89#
b.. #2x = - 81^@78 + k360^@#, or
#2x = 278^@22 + k360^@# (co-terminal)
#x = 139^@11 + k180^@#
The 2 smallest positive answers are:
#40^@89; 139^@11#