Question

How do you solve `cosx+ cos(3x) =0`?

Answer 1

`x = pi/2, (3pi)/2, pi/4, (3pi)/4, (5pi)/4 and (7pi)/4`

Explanation:

Note that `cos3x` can be rewritten as `cos(2x + x)`.

`cosx + cos(2x + x) = 0`

Now use `cos(A + B) = cosAcosB - sinAsinB`.

`cosx + cos2xcosx - sin2xsinx = 0`

Apply `cos2x = 2cos^2x -1` and `sin2x = 2sinxcosx`.

`cosx + (2cos^2x - 1)cosx - 2sinxcosx(sinx) = 0`

`cosx + 2cos^3x - cosx - 2sin^2xcosx = 0`

Use `sin^2x + cos^2x = 1`:

`cosx + 2cos^3x - cosx - 2(1 - cos^2x)cosx = 0`

`cosx + 2cos^3x - cosx - 2cosx + 2cos^3x = 0`

`4cos^3x - 2cosx = 0`

Factor:

`2cosx(2cos^2x - 1) = 0`

We have

`cosx = 0`

`x = pi/2, (3pi)/2`

AND

`cosx = +-1/sqrt(2)`

`x = pi/4, (3pi)/4, (5pi)/4 and (7pi)/4`

Hopefully this helps!

Answer 2

`Cosx+cos3x=0`

`=>2Cos2xcosx=0`

When `cosx=0=cos (pi/2)`

This the general solution as

`=>x=2npipmpi/2" where " n in ZZ`

To get the solution `" " x in [0.2pi]`

For n =0

`x=pi/2`

For n =1

`x=2pi-pi/2=(3pi)/2`

Again when `cos2x=0=cos (pi/2)`

This the general solution as

`=>2x=2npipmpi/2" where " n in ZZ`

`=>x=npipmpi/4" where " n in ZZ`

To get the solution `" " x in [0.2pi]`

For n =0

`x=pi/4`

For n =1

`x=pi+pi/4=(5pi)/4`

`x=pi-pi/4=(3pi)/4`

For n =2

`x=2pi-pi/4=(7pi)/4`