Another trigonometric equation problem

• Apr 30th 2011, 10:01 PM
Glitch
Another trigonometric equation problem
The question:
Find all solutions $z \in C$ of cos(z) = cos(2)

My attempt:
cos(z) = cos(x)cosh(y) - isin(x)sinh(y) = cos(2)

So we need:
sin(x)sinh(y) = 0 [1]
cos(x)cosh(y) = cos(2) [2]

[1] is zero when $x = k\pi$ or y = 0
Substitute into [2]:
$cos(k\pi)cosh(0) \ne cos(2)$

Try just y = 0:
So cos(x)cosh(0) = cos(x) which we want to equal cos(2), thus $x = \pm2$

Try just $x = k\pi$
So $cos(k\pi)cosh(y) = \pm cosh(y) = cos(2)$

Case when k is odd:
cosh(y) = -cos(2)

Case when k is even:
cosh(y) = cos(2)

Therefore, y = $cosh^-1(cos(2))$

So, $z = \pm 2 + icosh^{-1}(cos(2))$

But the solution is $\pm(2 + 2k\pi)$ :/

What am I doing wrong? Thanks.
• May 1st 2011, 12:45 AM
FernandoRevilla
Quote:

Originally Posted by Glitch
Try just $x = k\pi$ So $cos(k\pi)cosh(y) = \pm cosh(y) = cos(2)$

We have

$x=k\pi \Rightarrow \cos (k\pi)\cosh y=(-1)^k \cosh y$

But

$\cosh y \geq 1\; (\forall y\in\mathbb{R})\textrm{\;and\;}\cos 2\neq \pm 1$

So, we only obtain solutions for y=0.