help! adv math homework help!!!! plz?

**ericchung912** · May 27th 2008, 07:21 PM

square root of 3 times the cot times 3x plus one equals 0..

**Reckoner** · May 27th 2008, 07:38 PM

Originally Posted by ericchung912

square root of 3 times the cot times 3x plus one equals 0..

The cotangent of what? Do you mean $\sqrt3\cot\left(3x + 1\right) = 0$ ? Or maybe $\sqrt{3\cot\left(3x + 1\right)} = 0$ ? Or $\sqrt3\cot\left(3x\right) + 1 = 0$ ? $\sqrt{3\cot\left(3x\right)} + 1 = 0$ ? $\sqrt{3\cot\left(3x\right) + 1} = 0$ ? $\sqrt3\left[\cot\left(3x\right) + 1\right] = 0$ ?

**Chris L T521** · May 27th 2008, 07:45 PM

Originally Posted by ericchung912

square root of 3 times the cot times 3x plus one equals 0..

Is this what you mean?

$\sqrt{3}\cot(3x)+1=0$

Is there a restriction on x? (i.e. $0 \leq x <2\pi$ )?

After some modifications, we get:

$\cot(3x)=-\frac{1}{\sqrt{3}}$

Note that $\cot\theta=-\frac{1}{\sqrt{3}}$ when $\theta=\frac{2\pi}{3}\text{,} \ \frac{5\pi}{3}\text{,} \ \dots \ \frac{(3n-1)\pi}{3}; \ n\in\mathbb{N}$

Thus, $3x=\frac{2\pi}{3}\text{,} \ \frac{5\pi}{3}\text{,} \ \dots \ \frac{(3n-1)\pi}{3}; \ n\in\mathbb{N}$
$\implies x=\frac{2\pi}{9}\text{,} \ \frac{5\pi}{9}\text{,} \ \dots \ \frac{(3n-1)\pi}{3}; \ n\in\mathbb{N}$ .

I hope this helped (assuming that I interpreted the question properly)!

**ericchung912** · May 27th 2008, 07:48 PM

thats what i mean ypu.. knda new. here so i dont know how to use it
$ \sqrt{3}\cot(3x)+1=0 $

the directions to this problem is:
solve the following equations that o is less than or equal to x...x is less than 360

**Chris L T521** · May 27th 2008, 07:59 PM

Originally Posted by ericchung912

thats what i mean ypu.. knda new. here so i dont know how to use it
$ \sqrt{3}\cot(3x)+1=0 $

the directions to this problem is:
solve the following equations that o is less than or equal to x...x is less than 360

Ok then! Referring back to my solution (I had it in radians, now I'm gonna convert it to degrees...), we saw that:

$x=\frac{2(360^o)}{9}\text{,} \ \frac{5(360^o)}{9}\text{,} \ \frac{8(360^o)}{9}\text{,} \ \frac{11(360^o)}{9}\text{,} \ \frac{14(360^o)}{9}\text{,} \ \dots \ \frac{(3n-1)(360^o)}{9}; \ n\in\mathbb{N}$

After simplifying we get:

$\color{red}\boxed{x=80^o\text{,} \ 200^o\text{,} \ 320^o}$

The ones after 320 are outside of the domain ( $0\leq x<360^o$ ).

Hope this made sense!

**Chris L T521** · May 27th 2008, 08:16 PM

Originally Posted by Chris L T521

Ok then! Referring back to my solution (I had it in radians, now I'm gonna convert it to degrees...), we saw that:

$x=\frac{2(360^o)}{9}\text{,} \ \frac{5(360^o)}{9}\text{,} \ \frac{8(360^o)}{9}\text{,} \ \frac{11(360^o)}{9}\text{,} \ \frac{14(360^o)}{9}\text{,} \ \dots \ \frac{(3n-1)(360^o)}{9}; \ n\in\mathbb{N}$

After simplifying we get:

$\color{red}\boxed{x=80^o\text{,} \ 200^o\text{,} \ 320^o}$

The ones after 320 are outside of the domain ( $0\leq x<360^o$ ).

Hope this made sense!

Forgive me, but I made a "little" error

$\pi = 180^o\neq 360^o$

Thus, $\color{red}\boxed{x=40^o\text{,} \ 100^o\text{,} \ 160^o\text{,} \ 220^o\text{,} \ 280^o\text{,} \ 340^o}$

To get to the new answers, divide the old ones by 2 (you will have more answers...I hope you can follow).

Sorry about that!

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