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

1 Answer

Apr 8, 2018

#x=(7pi)/6+2pin, x=(11pi)/6+2pin#

Explanation:

Examine the unit circle to see where the sine is #-1/2.#

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#x=(7pi)/6, x=(11pi)/6# give the desired value. But sine is periodic, all values of sine repeat every #2pi# units.

To account for that we'll add #2pin# to each answer where #n# is any integer.

#x=(7pi)/6+2pin, x=(11pi)/6+2pin#

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