Question

How do you solve the identity `cos 2x + cos 4x = 0`?

Answer 1

Solve f(x) = cos 2x + cos 4x = 0

Explanation:

Apply the trig identity:`cos 4x = 2cos^2 2x - 1.`
`f(x) = cos 2x + 2cos^2 2x - 1 = 0`
Call cos 2x = t, we have to solve the quadratic equation:

2t^2 + t - 1 = 0

Since (a - b + c) = 0, the shortcut gives: `t = - 1` and `t = -c/a = 1/2`

a. `cos 2x = t = -1`--> `2x = +- pi` --> `x = +- pi/2`
b. `cos 2x = t = 1/2` --> `2x = +- pi/3` --> `x = +- pi/6`
Within interval `(-pi, pi),` there are 6 answers:
`+- pi/2 ; +- pi/3 and +- pi/6`
Check by calculator:
`x = pi/2` -> cos 2x = cos pi = -1 ; cos 4x = cos 2pi = 1 -->
--> f(x) = -1 + 1 = 0. OK
`x = pi/6` --> `cos 2x = cos pi/3 = 1/2` ; `cos 4x = cos (2pi)/3 = -1/2`
-->`f(x) = 1/2 - 1/2 = 0.` OK
`x = pi/3` --> cos 2x = cos (2pi)/3 = 1/2 ; cos 4x = cos (4pi)/3 = - 1/2
f(x) = 1/2 - 1/2 = 0. OK

Answer 2

`color(blue)(x= pi/6+npi)color(white)("XX")orcolor(white)("XX")color(blue)(x = (5pi)/6+npi)color(white)("XX")orcolor(white)("XX")color(blue)(x =pi/2+npi)`
`color(white)("XXXXXXXX")AA n in ZZ`

Explanation:

Let `theta = 2x`

`cos(2x)+cos(4x)=0`
`color(white)("XXXX")`becomes
`cos(theta)+cos(2theta) = 0`

By double angle formula
`color(white)("XXXX")cos(2theta) = 2cos^2(theta)-1`

So
`cos(theta)+(2cos^2(theta) -1) = 0`

Rearranging
`2cos^2(theta)+cos(theta)-1 = 0`

Factoring
`(2cos(theta)-1)(cos(theta)+1) = 0`

`rArrcolor(red)(cos(theta) = 1/2)color(white)("XXXXXXXXXXXX")orcolor(white)("XXX")color(blue)(cos(theta)=-1)`

if restricted to `theta in [0,2pi]`
`color(red)(theta = pi/3 or theta=(5pi)/3)color(white)("XXXXXXXXXXX")orcolor(white)("XXXX")color(blue)(theta=pi)`

or in general (with `n in ZZ`)
`color(red)(theta = pi/3+n2pi or theta=(5pi)/3+n2pi)color(white)("XXX")orcolor(white)("XXXX")color(blue)(theta=pi+2pi)`

Since `theta = 2x`
`x= pi/6+npi or x = (5pi)/6+npicolor(white)("XXXXX")orcolor(white)("XXXX")x =pi/2+npi`

Answer 3

`pi/2 + 2kpi`
`(3pi)/2 + 2kpi`
`pi/6 + (2kpi)/3`
`pi/2 + (2kpi/3)`

Explanation:

Apply trig identity:
`cos a + cos b = 2cos ((a + b)/2)cos ((a - b)/2)`
In this case:
`cos 2x + cos 4x = 2cos 3x.cos x = 0`
Either cos x or cos 3x should be zero.

a. cos x = 0 --> Unit circle gives 2 solutions:
`x = pi/2 + 2kpi`, and `x = (3pi)/2 + 2kpi`
b/ cos 3x = 0 --> unit circle gives 2 solutions:
`3x = pi/2 + 2kpi` --> `x = pi/6 + (2kpi)/3`
`3x = (3pi/2) + 2kpi` --> `x = pi/2 + (2kpi)/3`