How do you find all solutions to the equation #sin x=1/2#, when x is less than equal to 0 and less than equal to 2pi?
1 Answer
Jun 3, 2018
Explanation:
#"since "sinx>0" then x is in first/second"#
#"quadrants"#
#x=sin^-1(1/2)=pi/6larrcolor(red)"in first quadrant"#
#"or "x=(pi-pi/6)=(5pi)/6larrcolor(red)"second quadrant"#
#"solutions are "x=pi/6" or "x=(5pi)/6to(0<=x<=2pi)#
