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

3 Answers
Apr 14, 2018

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

Apr 14, 2018

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

Apr 14, 2018

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)