Question

How do you solve for x in `2cos(3x-1)=0`?

Answer 1

If `2cos(3x-1)=0` then so is `cos(3x-1)`

If a `cos theta=0` then `theta=90^0` plus any multiple of `180^0`
(the cosine function crosses the `x`-axis twice every period)

So we have
`3x-1=90^0+k*180^0`

`->3x=91^0+k*180^0`

`->x=91^0/3+k*180^0/3`

Answer:
`x=(30 1/3) ^0 +k*60^0`

Extra:
If you work in radians (you didn't say), then solving would go along the same lines, only then you would work with

`3x-1=pi/2 +k.pi`