Question

How do you solve `2 sin x - 1 = 0` over the interval 0 to 2pi?

Answer 1

`x = pi/6 , 5pi/6`

Explanation:

1/ `2sin(x) - 1 = 0`
2/ `2sin(x) = 1`
3/ `sin(x) = 1/2`

4/ `x = pi/6 , 5pi/6`

Answer 2

`x=pi/6 or (5pi)/6`

Explanation:

`2sin(x)-1=0|+1`
`2sin(x)=1|:2`
`sin(x)=1/2`
`x=arcsin(1/2)=pi/6 or (5pi)/6`

Answer 3

`x=pi/6,(5pi)/6`

Explanation:

`2sinx-1=0`

`rArrsinx=1/2`

`"since "sinx>0" then x in first/second quadrant"`

`rArrx=sin^-1(1/2)=pi/6larrcolor(blue)"first quadrant"`

`"or "x=pi-pi/6=(5pi)/6larrcolor(blue)"second quadrant"`

`rArrx=pi/6,(5pi)/6to(0,2pi)`