Author Topic: Freds spech qns (Read 1512 times) Tweet Share

fredrick · « **on:** October 20, 2008, 07:11:28 pm »

$\theta\in[-2\pi,2\pi]$ , find all solutions to the equation $cosec(\theta)+cosec(\frac{\theta}{2})=0$
Heres my working
$\frac{1}{sin(\theta)} + \frac{1}{sin(\frac{\theta}{2})}=0$
$sin({\frac{\theta}{2})} + sin(\theta)=0$

$sin({\frac{\theta}{2})} + 2sin({\frac{\theta}{2})}cos({\frac{\theta}{2})} =0$

$[sin({\frac{\theta}{2})} ][1+2cos({\frac{\theta}{2})}]=0$

$sin({\frac{\theta}{2})}=0$ and $cos({\frac{\theta}{2})}=-\frac{1}{2}$ , where $\frac{\theta}{2}\in[-\pi,\pi]$
So $\theta=-2\pi,-\frac{4\pi}{3},0,\frac{4\pi}{3},2\pi$

But the solutions only have $\theta=-\frac{4\pi}{3},\frac{4\pi}{3}$

Collin Li · « **Reply #1 on:** October 20, 2008, 07:19:31 pm »

This is because $\sin{\left(\frac{\theta}{2}\right)} \neq 0$ by virtue of the fact that the equation contains $\mbox{cosec}{\left(\frac{\theta}{2}\right)} = \frac{1}{\sin{\left(\frac{\theta}{2}\right)}}$

Plugging any of $\theta = -2\pi,\, 0,\, 2\pi$ values into the original equation would cause division by zero.

fredrick · « **Reply #2 on:** October 20, 2008, 07:43:39 pm »

sweet thxs

Pandemonium · « **Reply #3 on:** October 21, 2008, 10:41:25 pm »

Quote from: coblin on October 20, 2008, 07:19:31 pm

This is because $\sin{\left(\frac{\theta}{2}\right)} \neq 0$ by virtue of the fact that the equation contains $\mbox{cosec}{\left(\frac{\theta}{2}\right)} = \frac{1}{\sin{\left(\frac{\theta}{2}\right)}}$

Plugging any of $\theta = -2\pi,\, 0,\, 2\pi$ values into the original equation would cause division by zero.

Quote from: coblin on October 20, 2008, 07:19:31 pm

Plugging any of $\theta = -2\pi,\, 0,\, 2\pi$ values into the original equation would cause division by zero.

Quote from: coblin on October 20, 2008, 07:19:31 pm

values into the original equation would cause division by zero.

Quote from: coblin on October 20, 2008, 07:19:31 pm

equation would cause division by zero.

Quote from: coblin on October 20, 2008, 07:19:31 pm

cause division by zero.

Quote from: coblin on October 20, 2008, 07:19:31 pm

division by zero.

OH SHIIIIT THE WORLD IS GOING TO END. EVERYONE GET UNDER A TABLE NOW!

Flaming_Arrow · « **Reply #4 on:** October 21, 2008, 10:42:48 pm »

EPIC LULZ

fredrick · « **Reply #5 on:** October 22, 2008, 03:06:33 pm »

Another question:
$\int\frac{x}{3+x}dx$

I got $3+x-3log|3+x|+c$

The answer is $x-3log|3+x|+c_{2}$ , where $c_{2}=3+c_{1}$
i know its essentially the same thing but will you still get full marks if you leave it in the first form?

peace

riadnicolas · « **Reply #6 on:** October 22, 2008, 04:33:36 pm »

Quote from: fredrick on October 22, 2008, 03:06:33 pm

Another question:
$\int\frac{x}{3+x}dx$

I got $3+x-3log|3+x|+c$

The answer is $x-3log|3+x|+c_{2}$ , where $c_{2}=3+c_{1}$
i know its essentially the same thing but will you still get full marks if you leave it in the first form?

peace

yeh, i also wanted to know is it wrong to do such a question by long division rather than u sub

Collin Li · « **Reply #7 on:** October 22, 2008, 04:55:13 pm »

No it isn't wrong, because the constant of integration makes the other constants in the expression irrelevant.

However, if you have an incompetent examiner, they might mark you wrong.

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