Question

How do you evaluate `Arccos (- sqrt 3 / 2)`?

Answer 1

`(5pi)/6 and (7pi)/6`

Explanation:

Trig table and unit circle --> `cos x = (-sqrt3/2)` -->
arc `x = +- (5pi)/6`
Answers for `(0, 2pi)`;
`(5pi)/6 and (7pi)/6` (co-terminal to `-(5pi)/6`)