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How do you solve #Cos6x=sqrt3/2#?

↳Redirected from "How do you evaluate #sin^-1 (-sqrt3/2)#?"

1 Answer

Aug 1, 2016

#x=5#

Explanation:

#cos6x=sqrt3/2#
or
#cos6x=cos30#
or
#6x=30#
or
#x=30/6#
or
#x=5#

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