How do you solve #sin(x)=1/2#?

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How do you solve #sin(x)=1/2#?

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Answer 1 · 2015-04-14T20:21:02

The solutions are:

#x=pi/6+2kpi#

and

#x=5/6pi+2kpi#.

Draw the trigonometric circle and a line passing from #O# with a slope of #alpha=pi/6#. That line intercept the circle in a point #A#, make the perpendicular from that point to the x-axis, the interception is #B#.
The triangle #OAB# is half of an equilater triangle with #OA=1# and #AB=1/2#. So #sinalpha=(AB)/(OA)=1/2#.

You can say the same with the other line with #alpha=5/6pi#.

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How do you solve #sin(x)=1/2#?

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