How do you find the value for #cos^ -1 (- sqrt 3/2)#?

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Sep 23, 2015

Answer:

#5pi/6#radians or #150^o#

Explanation:

as #-sqrt3/2 is Negative#
and cos function is negative in 2nd and 3rd quadrant
#-sqrt3/2=cos(pi-pi/6)or cos(pi+pi/6)#
as #arccosx# range is in#[0 , pi]##cos(pi+pi/6) #is ruled out
#-sqrt3/2=cos(pi-pi/6)##=>-sqrt3/2= arccoscos(pi-pi/3)=pi-pi/6=5pi/6radians =5/6*180^o=150^o#

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