Question

How do you solve `cos(t)= -sqrt(3)/2` over the interval 0 to 2pi?

Answer 1

`t=(5pi)/6=150^@` and also `t=(7pi)/6=210^@`

Explanation:

in the interval `0` to `2pi`, cos t is negative at the 2nd and 3rd quadrants

the reference angle is the special angle `30^@`

the angle `150^@` has a cosine of `-sqrt3/2` also
the angle `210^@` has a cosing of `-sqrt3/2`

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