# Thread: Find the value of x.

1. ## Find the value of x.

For $\displaystyle \pi<x<\frac{3\pi}{2}$ with $\displaystyle cos(x)=-sin(\frac{\pi}{6})$, find the value of $\displaystyle x$ (do not evaluate $\displaystyle sin(\frac{\pi}{6})$).

Please show all steps to any working out.

Any help will be appreciated!

2. Originally Posted by Joker37
For $\displaystyle \pi<x<\frac{3\pi}{2}$ with $\displaystyle cos(x)=-sin(\frac{\pi}{6})$, find the value of $\displaystyle x$ (do not evaluate $\displaystyle sin(\frac{\pi}{6})$).

Please show all steps to any working out.

Any help will be appreciated!
Notice that $\displaystyle -sin(\frac{\pi}{6}) = sin(\frac{-\pi}{6})$

3. ## http://latex.codecogs.com/png.latex?%5Cpi%3Cx%3C%5Cfrac{3%5Cpi}{2}

Originally Posted by Joker37
For $\displaystyle \pi<x<\frac{3\pi}{2}$ with $\displaystyle cos(x)=-sin(\frac{\pi}{6})$, find the value of $\displaystyle x$ (do not evaluate $\displaystyle sin(\frac{\pi}{6})$).

Please show all steps to any working out.

Any help will be appreciated!
$\displaystyle sin(\pi /6)=\frac{1}{2}$, so $\displaystyle -sin(\pi /6)=-\frac{1}{2}$, so we have:
cos(x)=-$\displaystyle \frac{1}{2}$
x=$\displaystyle \pm acos(-\frac{1}{2})+\pi n,n\in Z$
x=$\displaystyle \pm \frac{2\pi }{3} +\pi n,n\in Z$

if $\displaystyle \pi<x<\frac{3\pi}{2}$ that $\displaystyle x=\frac{4\pi}{3}$, or $\displaystyle x=-\frac{3\pi}{2}$

4. Originally Posted by ihavenonick
sin(π/6)=\frac{1}{2}, so -sin(п/6)=\frac-{1}{2}, so we have:
cos(x)=\frac-{1}{2}
x=\pm acos(-1/2)+πn,n\in Z
x=\pm 2π/3 +πn,n\in Z

if $\displaystyle \pi<x<\frac{3\pi}{2}$ that x=\frac-{3\pi}{2}, or x=\frac-{3\pi}{2}
Besides doing the entire problem, you evaluated sin(π/6) - something that was expressly forbidden!

5. Originally Posted by TheChaz
Notice that $\displaystyle -sin(\frac{\pi}{6}) = sin(\frac{-\pi}{6})$
sure thing!

6. Originally Posted by TheChaz
Besides doing the entire problem, you evaluated sin(π/6) - something that was expressly forbidden!
I was editing the post at that moment. Look at this now.

7. Originally Posted by ihavenonick
$\displaystyle sin(\pi /6)=\frac{1}{2}$, so $\displaystyle -sin(\pi /6)=-\frac{1}{2}$, so we have:
cos(x)=-$\displaystyle \frac{1}{2}$
x=$\displaystyle \pm acos(-\frac{1}{2})+\pi n,n\in Z$
x=$\displaystyle \pm \frac{2\pi }{3} +\pi n,n\in Z$

if $\displaystyle \pi<x<\frac{3\pi}{2}$ that $\displaystyle x=\frac{4\pi}{3}$, or $\displaystyle x=-\frac{3\pi}{2}$
Actually the answer is meant to be $\displaystyle \frac{7\pi}{6}$ but nevermind I think I understand how to do it now.

8. Originally Posted by ihavenonick
$\displaystyle sin(\pi /6)=\frac{1}{2}$, so $\displaystyle -sin(\pi /6)=-\frac{1}{2}$, so we have:
cos(x)=-$\displaystyle \frac{1}{2}$
x=$\displaystyle \pm acos(-\frac{1}{2})+\pi n,n\in Z$
x=$\displaystyle \pm \frac{2\pi }{3} +\pi n,n\in Z$

if $\displaystyle \pi<x<\frac{3\pi}{2}$ that $\displaystyle x=\frac{4\pi}{3}$, or $\displaystyle x=-\frac{3\pi}{2}$
Im sorry Ive done a mistake. Here is a correct answer:
$\displaystyle sin(\pi /6)=\frac{1}{2}$, so $\displaystyle -sin(\pi /6)=-\frac{1}{2}$, so we have:
cos(x)=-$\displaystyle \frac{1}{2}$
x=$\displaystyle \pm acos(-\frac{1}{2})+\pi n,n\in Z$
x=$\displaystyle \pm \frac{2\pi }{3} +\pi n,n\in Z$

if $\displaystyle \pi<x<\frac{3\pi}{2}$ that $\displaystyle x=\frac{4\pi}{3}$(or $\displaystyle x=-\frac{2\pi}{3}$)

9. Jet another:
cos(x)=-$\displaystyle -sin(\pi /6)$
x=$\displaystyle \pm acos(-sin(\pi /6))+\pi n,n\in Z$;$\displaystyle cos(3\pi /2 - \alpha )=-\sin \alpha$
x=$\displaystyle \pm acos(cos(3\pi /2 - \pi /6)+\pi n,n\in Z$
x=$\displaystyle \pm acos(cos(9\pi /6-\pi /6)+\pi n,n\in Z$
x=$\displaystyle \pm acos(cos(8\pi /6)+\pi n,n\in Z$
x=$\displaystyle \pm acos(cos(4\pi /3)+\pi n,n\in Z$
x=$\displaystyle \pm 4\pi /3 +\pi n,n\in Z$

if $\displaystyle \pi<x<\frac{3\pi}{2}$ that $\displaystyle x=\frac{4\pi}{3}$

P.S. I don`t understand too. If calculate, the correct answer is $\displaystyle \frac{7\pi}{6}$ as you wrote.