Question

How do you solve `sin3x=cos3x`?

Answer 1

Use `tan 3x = (sin 3x) / (cos 3x) = 1` to find:

`x = pi/12 + (n pi) / 3`

Explanation:

Let `t = 3x`

If `sin t = cos t` then `tan t = sin t / cos t = 1`

So `t = arctan 1 + n pi = pi/4+n pi` for any `n in ZZ`

So `x = t/3 = (pi/4 + n pi) / 3 = pi/12 + (n pi) / 3`

Answer 2

Solve sin 3x = cos 3x

Answer: `x = pi/12 + Kpi/3`

Explanation:

Use the complementary arcs relationship:`cos x = sin (pi/2 - x)`

`sin 3x = sin (pi/2 - 3x)`
a. `3x = pi/2 - 3x` + 2Kpi -> `6x = pi/2 +2Kpi ->`
`x = pi/12 + Kpi/3`

Within interval`(0,2pi)` there are 6 answers: `pi/12; (5pi)/12; (9pi)/12; (13pi)/12; (17pi)/12; and (21pi)/12.`

b. `3x = pi - (pi/2 - 3x) = pi/2 + 3x.` This equation is undefined.

Check
`x = pi/12 --> sin 3x = sin pi/4 = sqrt2/2`
`x = pi/12 --> cos 3x = cos pi/4 = sqrt2/2`
Therefore sin 3x = cos 3x:
You may check other answers.

Answer 3

`x={(pi/12+(2pik)/3),(" "color(black)and),(-pi/4 + (2pik)/3):}`

`kinZZ`

Explanation:

Here's another method which has its own uses.

First, send every thing to one side

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

Next, express `sin3x-cos3x` as `Rcos(3x+lambda)`

`R` is a positive real and `lambda` is an angle

`=>sin(3x)-cos(3x)= Rcos(3x+lambda)`

`=>-cos(3x)+sin(3x)= Rcos(3x)coslambda-Rsin(3x)sinlambda`

Equate the coefficients of `cosx` and `sinx` on both sides

`=>" "Rcoslambda=-1 " "...color(red)((1))`
`" "-Rsinlambda=1 " "...color(red)((2))`

`color(red)(((2))/((1)))=>-(-Rsinlambda)/(Rcoslambda)=1/(-1)`

`=>tanlambda=1=>lambda=pi/4`

`color(red)((1)^2) + color(red)((2)^2) => (Rcoslambda)^2+(-Rsinlambda)^2=(-1)^2+(1)^2`

`=>R^2(cos^2lambda+sin^2lambda)=2`

`=>R^2(1)=2=>R=sqrt(2)`

So, `sin(3x)-cos(3x)=sqrt(2)cos(3x+pi/4)=0`

`=>cos(3x+pi/4)=0`

`=>3x+pi/4=+-pi/2+2pik`

Where `kinZZ`

Make `x` the subject

`=>x=+-pi/6-pi/12+2pik`

So we two sets of solutions :

`color(blue)(x={(pi/12+(2pik)/3),(" "color(black)and),(-pi/4 + (2pik)/3):})`

When `k=0=>x=pi/12+(2pi(0))/3=pi/12`

and `x=-pi/4+(2pi(0))/3=-pi/4`

When `k=1=>x=pi/12+(2pi)/3=(9pi)/12=(3pi)/4`
and `x=-pi/4+(2pi)/3=(5pi)/12`