# Another trigonometric equation problem

• Apr 30th 2011, 09:01 PM
Glitch
Another trigonometric equation problem
The question:
Find all solutions $\displaystyle z \in C$ 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 $\displaystyle x = k\pi$ or y = 0
Substitute into [2]:
$\displaystyle cos(k\pi)cosh(0) \ne cos(2)$

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

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

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

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

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

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

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

What am I doing wrong? Thanks.
• Apr 30th 2011, 11:45 PM
FernandoRevilla
Quote:

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

We have

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

But

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

So, we only obtain solutions for y=0.