Question #4dd67

1 Answer
Jim G.
May 15, 2017

#(5pi)/6, (7pi)/6#

Explanation:

#color(orange)"Reminder"#

#• secx=1/(cosx)#

#rArr(-sqrt3) . 1/(cosx)=2#

#rArrcosx=-(sqrt3)/2larr" to be solved"#

#["since ratio is negative then x is in second/third quadrant"]#

#x=cos^-1(sqrt3/2)=pi/6larrcolor(red)"related acute angle"#

#rArrx=(pi-pi/6)=(5pi)/6#

#rArrx=(pi+pi/6)=(7pi)/6#

#rArrx=(5pi)/6,(7pi)/6 ,x in[0,2pi]#

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