Given that #4<=y<=6#, find the value of y for which #2cos((2y)/3)+sqrt3=0#?

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Answer 1 · 2018-02-17T16:43:33

#color(blue)(y=(5pi)/4 , (7pi)/4)#

#2cos((2y)/3)+sqrt(3)=0#

#cos((2y)/3)=(-sqrt(3))/2#

Take arc cosine of both sides:

#arccos(cos((2y)/3))=arccos(-sqrt(3)/2)#

#(2y)/3=arccos(-sqrt(3)/2)#

#arccos(-sqrt(3)/2)=(5pi)/6 , (7pi)/6#

We try these two first, to see if we are in the given interval.

#(2y)/3=(5pi)/6=>y=(5pi)/4~~3.927#

#(2y)/3=(7pi)/6=>y=(7pi)/4~~5.498#

Any greater angles will not be in the given interval.

So:

#color(blue)(y=(5pi)/4 , (7pi)/4)#