Question

Solve the equation `cos^2x+cos^2(3x)=1`?

Answer 1

`x=(kpi+-pi/4)uu((kpi)/2+-pi/8)` where `k` is an integer.

Explanation:

`cos^2(3x)=1-cos^2x=sin^2x` so

`(cos(3x)+sin(x))(cos(3x)-sin(x))=0`

or

1) `cos(3x)=-sin(x)=sin(-x)=cos(pi/2-(-x))`

then

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

i.e. either `2x=2kpi+pi/2` and `x=kpi+pi/4`

or `4x=2kpi-pi/2` and `x=(kpi)/2-pi/8`

2) `cos(3x)-sin(x)=0`

or

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

so

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

i.e. either `4x=2kpi+pi/2` and `x=(kpi)/2+pi/8`

or `2x=2kpi-pi/2` and `x=kpi-pi/4`

so the solutions are

`x=(kpi+-pi/4)uu((kpi)/2+-pi/8)` where `k` is an integer.

Answer 2

`{x:x={:(pi/4+npi),(("-"pi)/8+(npi)/2),(pi/8+(npi)/2),(("-"pi)/4+npi):},ninZZ}` or equivalently `{x:x=(npi)/8,n!=4m,ninZZ,m inZZ}`

Explanation:

`cos^2x+cos^2(3x)=1`

Subtract `cos^2x` from both sides
`cos^2(3x)=1-cos^2x`

Using the Pythagorean Identity `sin^2theta+cos^2theta=1`,
`1-cos^2x=sin^2x`, therefore
`cos^2(3x)=sin^2x`

Subtract `sin^2x` from both sides
`cos^2(3x)-sin^2x=0`

Factor the left side using the difference of squares
`(cos(3x)+sinx)(cos(3x)-sinx)=0`

Time to find the value of `x`

Case 1: `cos(3x)+sinx=0`

Subtract `sinx` from both sides
`cos(3x)="-"sinx`

Since `"-"sintheta=sin("-"theta)`,
`"-"sinx=sin("-"x)`, therefore
`cos(3x)=sin("-"x)`

Since `sintheta=cos(pi/2-theta)` and `cos(theta)=cos(theta+2npi),ninZZ`,
`sin("-"x)=cos(pi/2+x+2npi)`, therefore
`cos(3x)=cos(pi/2+x+2npi)`

Since `costheta=cos("-"theta)`,
`3x=pi/2+x+2npi or 3x=("-"pi)/2-x-2npi`

Case 1.1: `3x=pi/2+x+2npi`

Subtract `x` from both sides
`2x=pi/2+2npi`

Divide both sides by `2`
`color(blue)(x=pi/4+npi,ninZZ)`

Case 1.2: `3x=("-"pi)/2-x-2npi`

Add `x` to both sides
`4x=("-"pi)/2-2npi`

Divide both sides by `4`
`color(blue)(x=("-"pi)/8-(npi)/2,ninZZ)`

Case 2: `cos(3x)-sinx=0`

Add `sinx` to both sides
`cos(3x)=sinx`

Since `sintheta=cos(pi/2-theta)` and `cos(theta)=cos(theta+2npi),ninZZ`,
`sinx=cos(pi/2-x+2npi)`, therefore
`cos(3x)=cos(pi/2-x+2npi)`

Since `costheta=cos("-"theta)`,
`3x=pi/2-x+2npi or 3x=("-"pi)/2+x-2npi`

Case 2.1: `3x=pi/2-x+2npi`

Add `x` to both sides
`4x=pi/2+2npi`

Divide both sides by `4`
`color(blue)(x=pi/8+(npi)/2,ninZZ)`

Case 2.2: `3x=("-"pi)/2+x-2npi`

Subtract `x` from both sides
`2x=("-"pi)/2-2npi`

Divide both sides by `2`
`color(blue)(x=("-"pi)/4-npi,ninZZ)`

Since `n` can be any integer, any term with `"-"n` can be replaced with `n`, so compiling everything gives
`color(red)({x:x={:(pi/4+npi),(("-"pi)/8+(npi)/2),(pi/8+(npi)/2),(("-"pi)/4+npi):},ninZZ})` or equivalently `color(red)({x:x=(npi)/8,n!=4m,ninZZ,m inZZ})`