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
Jim G.
Jun 3, 2018

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

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)

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