Question

cosx + cos3x = 0 Can anyone explain ?

Answer 1

\[\cos(3x)=\cos(2x+x)=\cos(2x)\cos(x)-\sin(2x)\sin(x) \\ =(\cos^2(x)-\sin^2(x))\cos(x)-2\sin(x)\cos(x)\sin(x) \\ =\cos^3(x)-\sin^2(x)\cos(x)-2\sin^2(x)\cos(x) \\ =\cos^3(x)-3\sin^2(x)\cos(x) \\ \\ \text{ so we have } \cos(x)+\cos(3x)=0 \\ \cos(x)+\cos^3(x)-3\sin^2(x)\cos(x)=0 \\ \cos(x)[ 1+\cos^2(x)-3\sin^2(x)]=0 \\ \cos(x)[1+1-\sin^2(x)-3\sin^2(x)]=0 \\ \cos(x)[2-4\sin^2(x)]=0\]

Answer 2

that should make things easier

Answer 3

set both factors equal to 0 and solve

Answer 4

\[\cos C+\cos D=2\cos \frac{ C+D }{ 2 }\cos \frac{ C-D }{ 2 }\]

Answer 5

that might make things faster :p

Answer 6

https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities