Question #55c95

2 Answers
Alan P.
Sep 18, 2016

In the range #x in [0,2pi]#
#color(white)("XXX")x=(2pi)/3 or (4pi)/3 or pi#

Explanation:

#2sin^2(x)-3cos(x)-3=0#

#rarr 2(1-cos^2(x))-3cos(x)-3=0color(white)("XXX")#(using Double Angle formula)

#rarr -2cos^2(x)-3cos(x)-1=0color(white)("XXX")#(simplifying)

#rarr (2cos(x)+1)(cos(x)+1)=0color(white)("XXX")#(factoring)

#rarr cos(x)=-1/2color(white)("XXX")#or#color(white)("XXX")cos(x)=-1#

For #x in [0,2pi]#
#color(white)("XXX")x= pi+-pi/3color(white)("XXXX")#or#color(white)("XXX")x=pi#

[Add #+n*2pi, n in ZZ# if you want not to restrict #x# to the range #[0,2pi]#

P dilip_k
Sep 18, 2016

#2sin^2x-3cosx-3=0#

#=>2-2cos^2x-3cosx-3=0#

#=>2cos^2x+3cosx+1=0#

#=>2cos^2x+2cosx+cosx+1=0#

#=>2cosx(cosx+1)+1(cosx+1)=0#

#=>(2cosx+1)(cosx+1)=0#

So

#2cosx+1=0#

#=>cosx=-1/2=cos((2pi)/3)#

#:.x=2npi+-(2pi)/3 " where "n in ZZ#

Again

#cosx+1=0#

#=>cosx=-1=cos(pi)#

#:.x=2npi+pi=(2n+1)pi " where "n in ZZ#

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