Question

What is the solution for `cos^nx-sin^nx=1` witn `n in NN^+`?

Answer 1

`x=2kpi, k=0,+-1, +-2, +-3,...`, sans odd k, when n is odd. In these omitted cases, x = (2k+1)pi/2, for odd k and odd n.

Explanation:

Let me solve `cos^n x+ sin^n x=1`, instead.

Squaring both sides,

`(cos^(2n)x+sin^(2n)x)+2cos^nxsin^nx`

`=(1)+2cos^nx sin^nx=1`

So,

#(sinx cos x)^n=0 to( sin (2x))^n=0 to sin (2x)=0 to 2x=kpi to x = k/2pi,

k=0, +-1, +-2, +-3, ...#. This includes extraneous solutions for odd k

against even n.#

With this clue, let us consider the given problem.

Rearranging as `cos^n x-1+sin^n x` and squaring,

`cos^(2n) x=(1+sin^n x)^2=sin^(2n) x+2 sin^n x+1`

Rearranging,

`1-2 sin^nx=(cos^(2n)x-sin^(2n) x)=1`

So,

`sin^nx=0 to sin x =0 to x=kpi, k=0,+-1, +-2, +-3,...`.

Here, odd k gives extraneous solution for odd n.

To include these cases as well, let us try squaring the alternative

rearrangement.

`sin^(2n) x=(cos^n x-1)^2 to cos^n x=0 to cos x =0 to x=(2k+1)pi/2`,

k=+-1, +-3, +-5.... n=1, 3, 5, ..#

Answer 2

If `n` is even the solution is `x=pm kpi, k=0,1,2,cdots`
If `n` is odd the solution is `x=-pi/2pm2kpi, k=0,1,2,cdots`

Explanation:

For `n=1` we have

`cosx-sinx=1` or `cosx+cos(x+pi/2)=1`

We know that

`cos(a+b)+cos(a-b)=2cosacosb` so using `a=x+pi/4, b=-pi/4` we have

`cosx-sinx=sqrt(2)cos(x+pi/4)=1` Solving for `x` we have

`x = 0+2kpi` and `x = -pi/2+2kpi` for `k=0,1,2,3,cdots`

Now supposing `n=2k` an even integer

`cos^(2k)x-sin^(2k)x = 1->cos^(2k)x=1+sin^(2k)x`

but `abs (cos x) le 1` so the solution is

`sinx=0` or `x=pm kpi, k = 0,1,2,cdots`

now supposing `n=2k+1` an odd integer, we have

`cos^(2k+1)x-sin^(2k+1)x = cos^(2k+1)(-x)-sin^(2k+1)(-x) = cos^(2k+1)(y)+sin^(2k+1)(y) = 1` with `y=-x`

but

`cos^2y cos^(2k-1)y-sin^2ysin^(2k-1)yle cos^2y+sin^2y = 1`

so the equality is observed only when

`cos^(2k-1)y=1, sin^(2k-1)y=0` or when `cos^(2k-1)y=0, sin^(2k-1)y=1`
and the solutions are

`y=kpi` for the first case and `y=pi/2+2kpi` for the second.

Ressuming

If `n` is even the solution is `x=pm kpi, k=0,1,2,cdots`
If `n` is odd the solution is `x=-pi/2pm2kpi, k=0,1,2,cdots`