Question

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.

Answer 1

`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`