Sin^4x -cos^4x= cos3x Could you solve this?

We have:

`(sin^2x+ cos^2x)(sin^2x- cos^2x) = cos(3x)`

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

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

`0 = cos(3x) + cos(2x)`

`0 = cos(2x)cos(x) - sin(2x)sinx + cos(2x)`

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

`0 = 2cos^3x - cosx - 2sin^2xcosx + 2cos^2x - 1`

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

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

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

`0 = 4cos^3x + 2cos^2x - 3cosx -1`

Let `u = cosx`.

`0 = 4u^3 + 2u^2 - 3u - 1`

We see that `u = -1` is a factor. Using synthetic division we get

`0 = (x + 1)(4x^2 - 2x - 1)`

The equation `4x^2 - 2x - 1= 0` may be solved using the quadratic formula.

`x = (2 +- sqrt(2^2 - 4 * 4 * -1))/(2 * 4)`

`x = (2 +- sqrt(20))/8`

`x = (1 +- sqrt(5))/4`

`x ~~ 0.809 or -0.309`

Since `cosx = u`, we get `x = pi/5, (3pi)/5` and `pi`.
Where `n` is an integer.

The graph of `y_1 = sin^4x- cos^4x` and `y_2 = cos(3x)` confirms that the solutions are the intersection points.

Hopefully this helps!

`sin^4x - cos^4 x = cos 3x`
`(sin^2 x + cos^2 x)(sin^2 x - cos^2 x) = cos 3x`
`1(sin^2 x - cos^2 x) = cos 3x`
`-cos 2x = cos 3x`, or
`cos 3x = - cos 2x = cos (2x + pi)`
Unit circle, and property of cos, give -->
`3x = +- (2x + pi) + 2kpi`
a. `3x = 2x + pi + 2kpi`
`x = (2k + 1)pi`
If k = 0 --> `x = pi`
b. `3x = - 2x - pi + 2kpi`
`5x = (2k - 1)pi`,
`x = ((2k - 1)pi)/5`
If k = 1 --> `x = pi/5`.
If k = 0 --> `x = - pi/5`, or `x = (9pi)/5` (co-terminal)
If k = 2 --> `x = (3pi)/5`
In the closed interval [0, 2pi], the answers are:
`0, (pi)/5, (3pi)/5, pi, (9pi)/5`
Check by calculator.
`x = pi/5 = 36^@` --> `sin^4 x = 0.119` --> `cos^4 x = - 0.428` --> cos 3x = - 309.
`sin^4 x - cos^4 x = 0.119 - 0.428 = - 309` . Proved
`x = (9pi)/5` -->`sin^4 x = 0.119` --> `cos^4 x = 0.428` -->
`sin^4 x - cos^4 x = - 0.309`
`cos 3x = cos 972 = - 0.309`. Proved

`rarrsin^4x-cos^4x=cos3x`

`rarr(sin^2x+cos^2x)(sin^2x-cos^2x)=cos3x`

`rarr-cos2x=cos3x`

`rarrcos3x+cos2x=0`

`rarr2cos((3x+2x)/2))*cos((3x-2x)/2))=0`

`rarrcos((5x)/2)*cos(x/2)=0`

Either `cos((5x)/2)=0`

`rarr(5x)/2=(2n+1)pi/2`

`rarrx=(2n+1)pi/5` `nrarrZ`

`rarrcos(x/2)=0`

`rarrx/2=(2n+1)pi/2`

`rarrx=(2n+1)pi` `nrarr`

I don't like reading other people's answers before I solve a question myself. But a featured answer for this one popped up. During my quick glance I couln't help notice it looked pretty complicated for what looks to me like a relatively easy question. I'll give it a shot.

`sin ^4 x - cos ^4 x = cos 3x`

`(sin ^2 x + cos ^2 x)(sin ^2 x - cos ^2 x ) = cos 3x`

`-cos 2x = cos 3x`

`cos (180^circ - 2x) = cos 3x`

I've been on Socratic for a couple of weeks, and this is emerging as my theme: The general solution to `cos x = cos a` is `x = \pm a + 360 ^circ k quad` for integer `k.`

`180^circ - 2x = \pm 3x + 360^circ k`

`-2x \pm 3x = -180^circ + 360^circ k`

We take the signs separately. Plus first:

`x = -180^circ + 360^circ k = 180 ^circ + 360^circ k`

Minus next.

`-5x = -180^circ + 360^circ k`

`x = 36^circ + 72^circ k`

If you read these closely you might think I'm making a mistake with the way I manipulate `k`. But since `k` ranges over all the integers, substitutions like `k to -k` and `k to k+1` are allowed and I slip those in to keep the signs `+` when they can be.

Check:

Let's pick a couple to check. I'm geeky enough to know `cos 36^circ` is half the Golden Ratio, but I'm not going to work these out exactly, just pop them into Wolfram Alpha to make sure.

`x = 36^circ + 72^circ = 108^circ`

`sin ^4 108 - cos ^4 108 - cos (3 * 108) = 0 quad sqrt`

`x = 180 - 2 (360) = -540`

`sin ^4 (-540) - cos^4(-540) - cos(3 * -540) = 0 quad sqrt`