I am doing a practice exam for my classes midterm and got stuck on these questions.

sinx=\(\frac{-\sqrt{3}}{2}\)

The answers that they provide for this question are 240° and 300°. I don't know how to get there though. The way I am solving this is by using x= inverse sine of \(\frac{-\sqrt{3}}{2}\) which gives me -60°. I dont know how to get to 240 nor 300.

cosx=\(\frac{\sqrt{2}}{2}\)

I have not tried this problem but given that it is the same section as the previous one, I still wanted help on it.

gbeans44 Jul 11, 2024