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

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

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Answer 1 · 2018-03-12T13:23:39

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

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

now you can notice a polynomial function by rearranging.

$4 {cos}^{3} (x) + 2 {cos}^{2} (x) - 2 cos (x) - 1 = - \frac{1}{2}$ $4Cos^3(x)+2cos^2(x)-2cos(x)-1=-1/2$

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

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

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

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