Question #60097

1 Answer
Jim G.
Feb 22, 2018

#x=(5pi)/6" or "x=(3pi)/2#

Explanation:

#2cos(x-pi/6)=-1#

#rArrcos(x-pi/6)=-1/2#

#rArrx-pi/6=cos^-1(1/2)=pi/3larrcolor(red)"related acute angle"#

#"since "cos(x-pi/6)<0#

#"then angle is in second/third quadrant"#

#rArrx-pi/6=pi-pi/3=(2pi)/3#

#"or "x-pi/6=pi+pi/3=(4pi)/3#

#rArrx=(2pi)/3+pi/6=(5pi)/6#

#"or "x=(4pi)/3+pi/6=(3pi)/2#

#rArrx=(5pi)/6,(3pi)/2in (0,2pi)#

Related questions
Impact of this question
198 views around the world
You can reuse this answer
Creative Commons License