Melody

$$\\(cos\frac{x}{2} )^2= 1\\\\
0= 1-(cos\frac{x}{2} )^2\\\\
(sin\frac{x}{2})^2=0\\\\
sin\frac{x}{2}=0\\\\
\frac{x}{2}=n\pi \qquad n\in Z \;\;$(N is an integer)$\\\\
x=2n\pi$$

I just realised that the question in your title is different from your question in your post  

$$\\x-2y=1\qquad (1a)\qquad x^2+y^2=13\qquad (2)\\
x=1+2y\\
x^2=1+4y^2+4y\qquad (1b)\\
sub\;1b\;into\;2\\
1+4y^2+4y+y^2=13\\
5y^2+4y-12=0\\
5y^2+10y-6y-12=0\\
5y(y+2)-6(y+2)=0\\
(5y-6)(y+2)=0\\

y=+6/5\;\;\;\;or\;\;\;\;y=-2$$

now sub thes y values back into 1a to find the corresponding x values.

you can finish it :)

Thanks Bertie, that is really neat.  It is so easy when YOU do it.   

s,k and t must all be positive integers I think.   :/

@@ End of Day Wrap  Thurs 25/6/15   Sydney, Australia Time   10:45pm  ♪ ♫

Hi everyone,

Our answer tributes for the past two days are bestowed on: Heureka, MathsGod1, Radix, Sir-Emo-Chappington, Nauseated, DarkBlaze347, Civoamzuk, Alan Anonymous4338, Hades, AydenSchnabel, TitaniumRome, Fiora and asinus.   Thank you  

Sun 28/6/15

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

Interest Posts: 

FTJ means 'For the juniors' 

Well Nauseated ....

"I might have thought so, but this forum is one of the few where I am not banned. I know because I can log on. Which begs the question: why do I want to be a member of a forum that would allow me as a member?

It’s not troll-like; it antithetical to our character --It’s a paradox. :)"

Yes - it is definitely a paradox      

Mind you - if you have been thrown out of all the other forum maybe you only want to be here because this is the ONLY place that will have you  :/

Plus, of course, we are also lovable.

Rosala is positively dripping with lovable ....      :) 

Thank you Alan and Sir-Emo, 

I intend to examine your answers in detail a little later.  

I am very curious about this one :)

[(p+1)-3] = p+1-3 = p-2

I am not sure what this question means   

$${\frac{\left({\mathtt{10}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}{\mathtt{\,\small\textbf+\,}}{\mathtt{29}}{\mathtt{\,\small\textbf+\,}}{\mathtt{32}}\right)}{{\mathtt{5}}}} = {\frac{{\mathtt{96}}}{{\mathtt{5}}}} = {\mathtt{19.2}}$$  

At the moment the average age is 19.2 years

Maybe it means that the average has to be between 13 and 25     

$$\\13<\frac{96+X}{6}<25\\\\
78<96+X<150\\\\
0

But really the person should be between  13 AND 25

I think that this is a dumb question  

You use inverse trig        OR   arc trig      (these are different terms for the same thing)

$$\\sin\theta =0.125\\\\
then\\\\
\theta = sin^{-1}(0.125) = asin(0.125)$$

On the web2 calc it is asin