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

• May 27th 2008, 07:21 PM
ericchung912
help! adv math homework help!!!! plz?
square root of 3 times the cot times 3x plus one equals 0..
• May 27th 2008, 07:38 PM
Reckoner
Quote:

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 $\sqrt3\cot\left(3x\right) + 1 = 0$? $\sqrt{3\cot\left(3x\right)} + 1 = 0$? $\sqrt{3\cot\left(3x\right) + 1} = 0$? $\sqrt3\left[\cot\left(3x\right) + 1\right] = 0$?
• May 27th 2008, 07:45 PM
Chris L T521
Quote:

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)! :D
• May 27th 2008, 07:48 PM
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
• May 27th 2008, 07:59 PM
Chris L T521
Quote:

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 ( $0\leq x<360^o$).

• May 27th 2008, 08:16 PM
Chris L T521
Quote:

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 ( $0\leq x<360^o$).

$\pi = 180^o\neq 360^o$
Thus, $\color{red}\boxed{x=40^o\text{,} \ 100^o\text{,} \ 160^o\text{,} \ 220^o\text{,} \ 280^o\text{,} \ 340^o}$