Question

How do you solve `2sin3x-1=0`?

Answer 1

First, move the `-1` to the other side (it will be `+1`):

`2sin3x=1`

Divide by 2:

`sin3x=1/2`

This belongs to an angle of `30^0` or `pi/6` radians

(plus any multiple of `360^0` or `2pi` radians)

So `3x=30^0+k.360^0->x=10^0+k*120^0`

(or `3x=pi/6 +k.2pi->x=pi/18 +k*2/3 pi`)