Trig help please?

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1 Answer
maganbhai P.
Feb 28, 2018

#(3) 120^0#

Explanation:

#2cos^2x-3cosx-2=0rArr2cos^2x-4cosx+cosx-2=0rArr2cosx(cosx-2)+1(cosx-2)=0rArr(cosx-2)(2cosx+1)=0rArrcosx=2 or cosx=-1/2#
But,#2!in[-1.1],#
So, #cosx=-1/2in[-1,1]#
#cosx=-1/2<0rArrcosx=cos(pi-pi/3)=cos((2pi)/3)#
#rArrx=(2pi)/3=120^0#

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