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

Nov 6, 2016

$x = 2 k \pi , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$, 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,

$\left({\cos}^{2 n} x + {\sin}^{2 n} x\right) + 2 {\cos}^{n} x {\sin}^{n} x$

$= \left(1\right) + 2 {\cos}^{n} x {\sin}^{n} x = 1$

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,

${\cos}^{2 n} x = {\left(1 + {\sin}^{n} x\right)}^{2} = {\sin}^{2 n} x + 2 {\sin}^{n} x + 1$

Rearranging,

$1 - 2 {\sin}^{n} x = \left({\cos}^{2 n} x - {\sin}^{2 n} x\right) = 1$

So,

${\sin}^{n} x = 0 \to \sin x = 0 \to x = k \pi , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$.

Here, odd k gives extraneous solution for odd n.

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

rearrangement.

${\sin}^{2 n} x = {\left({\cos}^{n} x - 1\right)}^{2} \to {\cos}^{n} x = 0 \to \cos x = 0 \to x = \left(2 k + 1\right) \frac{\pi}{2}$,

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

Nov 8, 2016

If $n$ is even the solution is $x = \pm k \pi , k = 0 , 1 , 2 , \cdots$
If $n$ is odd the solution is $x = - \frac{\pi}{2} \pm 2 k \pi , k = 0 , 1 , 2 , \cdots$

#### Explanation:

For $n = 1$ we have

$\cos x - \sin x = 1$ or $\cos x + \cos \left(x + \frac{\pi}{2}\right) = 1$

We know that

$\cos \left(a + b\right) + \cos \left(a - b\right) = 2 \cos a \cos b$ so using $a = x + \frac{\pi}{4} , b = - \frac{\pi}{4}$ we have

$\cos x - \sin x = \sqrt{2} \cos \left(x + \frac{\pi}{4}\right) = 1$ Solving for $x$ we have

$x = 0 + 2 k \pi$ and $x = - \frac{\pi}{2} + 2 k \pi$ for $k = 0 , 1 , 2 , 3 , \cdots$

Now supposing $n = 2 k$ an even integer

${\cos}^{2 k} x - {\sin}^{2 k} x = 1 \to {\cos}^{2 k} x = 1 + {\sin}^{2 k} x$

but $\left\mid \cos x \right\mid \le 1$ so the solution is

$\sin x = 0$ or $x = \pm k \pi , k = 0 , 1 , 2 , \cdots$

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

${\cos}^{2 k + 1} x - {\sin}^{2 k + 1} x = {\cos}^{2 k + 1} \left(- x\right) - {\sin}^{2 k + 1} \left(- x\right) = {\cos}^{2 k + 1} \left(y\right) + {\sin}^{2 k + 1} \left(y\right) = 1$ with $y = - x$

but

${\cos}^{2} y {\cos}^{2 k - 1} y - {\sin}^{2} y {\sin}^{2 k - 1} y \le {\cos}^{2} y + {\sin}^{2} y = 1$

so the equality is observed only when

${\cos}^{2 k - 1} y = 1 , {\sin}^{2 k - 1} y = 0$ or when ${\cos}^{2 k - 1} y = 0 , {\sin}^{2 k - 1} y = 1$
and the solutions are

$y = k \pi$ for the first case and $y = \frac{\pi}{2} + 2 k \pi$ for the second.

Ressuming

If $n$ is even the solution is $x = \pm k \pi , k = 0 , 1 , 2 , \cdots$
If $n$ is odd the solution is $x = - \frac{\pi}{2} \pm 2 k \pi , k = 0 , 1 , 2 , \cdots$