How do you solve cos 3x+cos 2x + cos x =-1/2 ?

1 Answer
peter k.
Mar 12, 2018

we need to use trigonometry identities such as the cosine triple angle identity which states, #Cos(3x)=4Cos^3(x)-3Cos(x)#
and the identity #cos(2x)=cos^2(x)-sin^2(x)#

then upon substituting this identities we have,
#4Cos^3(x)-3Cos(x)+cos^2(x)-sin^2(x)+cos(x)=-1/2#
but #sin^2(x)=1-cos^2(x)#
#4Cos^3(x)-3Cos(x)+cos^2(x)-1+cos^2(x)+cos(x)=-1/2#

now you can notice a polynomial function by rearranging.

#4Cos^3(x)+2cos^2(x)-2cos(x)-1=-1/2#

or#4Cos^3(x)+2cos^2(x)-2cos(x)=1/2#
now let cos(x)=a
#4a^3+2a^2-2a=1/2#
solving using a calculator gives #a=-0.22252,0.62349,-0.90097#

now #a=cos(x)=-0.2225 or 0.6235 or -0.9009#
solving the arc cosines of each gives the value's of #x# in rad.

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