How to find the solution of this equation?

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1 Answer
Mar 6, 2018

#color(red)(x=-pi/3,pi/3,(5pi)/3)#

Explanation:

we know that,#color(blue)(cos^-1(1/2)=cos^-1(cos(pi/3))=pi/3)#
Interval #color(blue)([-pi,2pi]=[-pi,0)uu[0,2pi])#.
So, #x=2npi+-pi/3,kinZ#
#rArrx=2npi-pi/3,kinZorx=2npi+pi/3,kinZ#
Taking #n=0.+-1.+-2,+-3,...#we get
#n=0rArrx=color(red)(-pi/3in[-pi,2pi]orx=pi/3in[-pi,2pi])#
#n=1rArrx=color(red)((5pi)/3in[-pi,2pi])orx=(7pi)/3!in[-pi,2pi]#
#n=2rArrx=(11pi)/3!in[-pi,2pi]orx=(13pi)/3!in[-pi,2pi]#
i.e.#AAninZ,n>=3rArrx!in[-pi,2pi]#
#n=-1rArrx=(-7pi)/3!in[-pi,2pi]orx=(-5pi)/3!in[-pi,2pi]#
i.e.#AAninZ,n<=-1rArrx!in[-pi,2pi]#
Thus, #color(red)(x=-pi/3,pi/3,(5pi)/3)#