#sin(x) > 1/2" in "[0,2pi)#?

How do I solve for #x#

1 Answer
Carl S.
Apr 2, 2018

#pi/6 < x < {5pi}/6#

Explanation:

Keep in mind that #sinx# is always greater than or equal to zero and less than or equal to 1.
#0 leq sinx leq 1#. That being said we have the following range: #1 geq sinx > 1/2#

Looking at the unit circle we can see that when #sinx =1/2#, #x=pi/6# or #x={5pi}/6#. This gives us a range of #pi/6 < x < {5pi}/6#

When #sinx=1#, #x = pi/2#
This is already within the range of #pi/6 < x < {5pi}/6# so we do not need to explicitly include that in our range.

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