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)#