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

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- May 27th 2008, 07:21 PMericchung912help! adv math homework help!!!! plz?
square root of 3 times the cot times 3x plus one equals 0..

- May 27th 2008, 07:38 PMReckoner
The cotangent of what? Do you mean $\displaystyle \sqrt3\cot\left(3x + 1\right) = 0$? Or maybe $\displaystyle \sqrt{3\cot\left(3x + 1\right)} = 0$? Or $\displaystyle \sqrt3\cot\left(3x\right) + 1 = 0$? $\displaystyle \sqrt{3\cot\left(3x\right)} + 1 = 0$? $\displaystyle \sqrt{3\cot\left(3x\right) + 1} = 0$? $\displaystyle \sqrt3\left[\cot\left(3x\right) + 1\right] = 0$?

- May 27th 2008, 07:45 PMChris L T521
Is this what you mean?

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

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

After some modifications, we get:

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

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

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

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

I hope this helped (assuming that I interpreted the question properly)! :D - May 27th 2008, 07:48 PMericchung912
thats what i mean ypu.. knda new. here so i dont know how to use it

$\displaystyle

\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 PMChris 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:

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

After simplifying we get:

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

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

Hope this made sense! :D - May 27th 2008, 08:16 PMChris L T521
Forgive me, but I made a "little" error (Headbang)

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

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

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! :D