Find the first four positive values of arccos(0.7). i know how to get the first, but as for the rest, i am lost.

Now cos⁻¹(0.7) is about 45.6°, that's on the first quadrant.

keep in mind that the inverse cosine function has a range of [0, 180°], so any angles it will spit out, will be on either the I quadrant where cosine is positive or the II quadrant, where cosine is negative.

however, 45.6° has a twin, she's at the IV quadrant, where cosine is also positive, and that'd be 360° - 45.6°, or 314.4°.

now, those are the first two, but we have been only working on the [0, 360°] range but we can simply go around the circle many times over up to 720° or 72000000000° if we so wish, so let's go just one more time around the circle to find the other fellows.

360° + 45.6° is a full circle and 45.6° more, that will give us the other angle, also in the first quadrant, but after a full cycle, at 405.6°.

then to find her twin on the IV quadrant, we simply keep on going, and that'd be at 360° + 360° - 45.6°, 674.4°.

and you can keep on going around the circle, but only four are needed this time only.