Question

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

Answer 1

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.