cos (t) = sqrt(3) /2 ... on the interval (0,2 pi)?

Answer 1

This would happen at t=pi/6 and at t=11pi/12

Answer 2

You need to do inverse trig to solve this.

if cos (t) = sqrt (3) /2, you need to find out what values of t, equal sqrt (3)/2 on the interval 0,2 pi, so basically, one complete circle.

t=arccos (sqrt 3)/2
cos is sqrt 3/2 at 30 degress or pi/6.

Now, in inverse trig, arccos is taken in Quadrant I and II. Because all trig functions are + in Quad I and cos is - in Quad II.

So our + answer for arccos is pi/6.
Our - answer for arccos in the II quadrant is, 5pi/6.