In a book of mine I stumbled upon this question and was wondering whether I have solved it properly.
The task is to determine all values of $t$ that makes the two vectors orthogonal
$u⃗=\binom{\cos(t)}{\sin(t)}$ and $x⃗=\binom{\sin(t)}{\cos(t)-2\sin(t)}$ where $t\in [0;2\pi]$
Here is my solution: We know that $u⃗\perp x⃗ \iff u⃗\cdot x⃗=0$
From that we have
$0=\cos(t)\cdot \sin(t)+\sin(t)\cdot (\cos(t)-2\sin(t))$
$0=2\cdot \sin(t)\cdot (\cos(t)-\sin(t))$
$0=\sin(t)\cdot (\cos(t)-\sin(t))$
That means we have:
$0=\sin(t)$ and $0=\cos(t)-\sin(t)$
Let's look at the first equation
$0=\sin(t)$ to this we can make a general solution: $\sin(n\pi)=0$ where $n\in [0;2\pi]$
From this we have 3 solutions;
$n=0,n=1,n=2$
This means that
$\sin(0)=0$ and $\sin(\pi)=0$ and $\sin(2\pi)=0$
For this reason our first 3 solutions are
$t=0,t=1,t=2$
Now we look at
$0=\cos(t)-\sin(t)$
$0={\cos(t)-\sin(t)\over \cos(t)}$
$0=1-\tan(t)$
$\tan(t)=1$
We know that
$\tan(45)=1$ because $\tan^{-1}(1)={\pi \over 4}$ this means that $\tan({\pi \over 4})=1$
Now we can say that if $\tan(t)=1$ has $t={\pi \over 4}$ as a solution then $n\pi+{\pi \over 4}$ is the general solution.
If we let $n=1$ we have $\pi+{\pi \over 4}={5\pi \over 4}$
To sum up all the solutions to this problem are:
$t=0,t=\pi ,t=2\pi ,t={\pi \over 4}$ and $t={5\pi \over 4}$
Basically, my questions are: have i solved this properly? and do any of you guys have a "cleaner" solution?