Find $x\in R$ such that undermentioned numbers create geometric sequence: $\cos(x), \sin(x), \tan(x)$. Find $x \in R$ such that undermentioned numbers create geometric sequence:
$\cos(x), \sin(x), \tan(x)$.
I know that $D:x \neq \frac{\pi}{2} + k\pi, k \in Z$.
I could get the result solving this equation:
$\frac{\sin(x)}{\cos(x)} = \frac{\tan(x)}{\sin(x)}$ but I'm not sure if I can do this because I don't know if $\sin(x) = 0$ or not. Or maybe I should do this for two cases?
EDIT:
Is this method correct?
As we can see in @Rory Daulton post we have: $\sin{x} = r\cos{x}, \tan{x} = r\sin{x}$. We can transform the first equation to the form: $r = \frac{\sin{x}}{\cos{x}} = \tan{x}$ (because due to the domain, $\cos{x} \neq 0$). So now we can transform the second equation to the form: $\tan{x} = \tan{x}\sin{x} \iff \tan{x}(\sin{x} - 1) = 0 \iff \tan{x} = 0 \vee \sin{x} = 1$.
 A: SHORT ANSWER: Yes, you can use cases, but you should use three cases. The first case is $\sin x=0$, the second is $\cos x=0$ (since that is also a denominator in your equation), and the third is your equation $\frac{\sin(x)}{\cos(x)} = \frac{\tan(x)}{\sin(x)}$. Solve those equations and check for extraneous solutions.
LONGER ANSWER: It is possible to avoid divisions in our solution method. In a geometric sequence of three elements we have $a_2=ra_1,\ a_3=ra_2$. So from your requirements we get
$$\sin x=r\cos x, \quad \tan x=r\sin x$$
Reversing the second equation and multiplying the two equations together we get
$$r\sin^2 x=r\tan x\cos x$$
Using $\tan x=\frac{\sin x}{\cos x}$, moving everything to the left, and factoring, we get
$$r\sin x (\sin x-1)=0$$
From that we get three cases:


*

*$r=0$


From the starting equations we get $\sin x=0,\ \tan x=0$. These have the same solutions $x=k\pi,\ k\in\Bbb Z$. Checking these answers we get the sequence
$$\pm 1,\ 0, 0$$
which is indeed a geometric sequence, with initial value $-1$ or $1$ and common ratio $0$.


*$\sin x=0$. This gives us the same solutions as case 1.

*$\sin x-1=0$. This has the solutions $x=2k\pi+\frac{\pi}2$. When we try check these answers we get the sequence
$$0,\ 1,\ \text{undefined}$$
which is not a geometric sequence for multiple reasons. So these answers are extraneous.
So the final answer is

$x=k\pi,\ k\in\Bbb Z$

