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

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

2. 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 $\sqrt3\cot\left(3x\right) + 1 = 0$? $\sqrt{3\cot\left(3x\right)} + 1 = 0$? $\sqrt{3\cot\left(3x\right) + 1} = 0$? $\sqrt3\left[\cot\left(3x\right) + 1\right] = 0$?

3. 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)!

4. 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

5. 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 ( $0\leq x<360^o$).

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