# cos(x -2pi/3) = 1/2

• Mar 20th 2014, 05:14 AM
andy000
cos(x -2pi/3) = 1/2
Hi,

the question I have been working on is cos(x -2pi/3) = 1/2 -2pi<x<2pi

I know from one of the special triangles that cos(pi/3) = 1/2 = cos(x-2pi/3)

so cos(x-2pi/3) = cos(pi/3), cos(-pi/3), cos(5pi/3), cos(-5pi/3), cos(7pi/3), cos(-7pi/3)

if this is right, should I just solve for each of them eg:

if cos(x-2pi/3) = cos(pi/3) ==> x-2pi/3 = pi/3 ==> x = 2pi/3+ pi/3 ==> x = pi

It would also be good if someone could let me know if I have missed any angles that give cos()= 1/2

thanks
• Mar 20th 2014, 06:32 AM
bjhopper
Re: cos(x -2pi/3) = 1/2
Quote:

Originally Posted by andy000
Hi,

the question I have been working on is cos(x -2pi/3) = 1/2 -2pi<x<2pi

I know from one of the special triangles that cos(pi/3) = 1/2 = cos(x-2pi/3)

so cos(x-2pi/3) = cos(pi/3), cos(-pi/3), cos(5pi/3), cos(-5pi/3), cos(7pi/3), cos(-7pi/3)

if this is right, should I just solve for each of them eg:

if cos(x-2pi/3) = cos(pi/3) ==> x-2pi/3 = pi/3 ==> x = 2pi/3+ pi/3 ==> x = pi

It would also be good if someone could let me know if I have missed any angles that give cos()= 1/2

thanks

cos 60 = 1/2
• Mar 20th 2014, 07:35 AM
SlipEternal
Re: cos(x -2pi/3) = 1/2
Quote:

Originally Posted by bjhopper
cos 60 = 1/2

That is not correct. $\cos 60^\circ = \dfrac{1}{2}$, but $\cos 60 \approx -0.95$.

Your value of $\dfrac{7\pi}{3}$ implies $x = 3\pi > 2\pi$, so that $x$ value is not in the domain. (But $\cos(x-2\pi/3) = \cos(-7\pi/3)$ does give an $x$-value in the domain).
• Mar 20th 2014, 08:33 AM
andy000
Re: cos(x -2pi/3) = 1/2
Hi,

so I guess I should have four values for x (discarding cos(7pi/3) and cos(5pi/3)?
• Mar 20th 2014, 08:35 AM
SlipEternal
Re: cos(x -2pi/3) = 1/2
Quote:

Originally Posted by andy000
Hi,

so I guess I should have four values for x (discarding cos(7pi/3) and cos(5pi/3)?

That is correct.
• Mar 20th 2014, 09:28 AM
andy000
Re: cos(x -2pi/3) = 1/2
Thanks again =)
• Mar 20th 2014, 01:11 PM
Soroban
Re: cos(x -2pi/3) = 1/2
Hello, andy000!

Quote:

$\text{Solve: }\:\cos\left(x -\tfrac{2\pi}{3}\right) \:=\:\tfrac{1}{2}\;\text{ for }-2\pi\,<\,x\,<\,2\pi$

We have: . $\cos\left(x - \tfrac{2\pi}{3}\right) \:=\:\tfrac{1}{2}$

. . $x - \tfrac{2\pi}{3} \;=\;\pm\tfrac{\pi}{3} + 2\pi n$

. . $x - \tfrac{2\pi}{3} \;=\;\begin{Bmatrix}\text{-}\frac{7\pi}{3} \\ \text{-}\frac{5\pi}{3} \\ \text{-}\frac{\pi}{3} \\ \frac{\pi}{3} \end{Bmatrix}$

. . $x \;=\;\begin{Bmatrix}\text{-}\frac{5\pi}{3} \\ \text{-}\pi \\ \frac{\pi}{3} \\ \pi \end{Bmatrix}$
• Mar 20th 2014, 07:37 PM
andy000
Re: cos(x -2pi/3) = 1/2
Hi Soroban,

Thanks for your reply. Can you explain how you get the equation x-2pi/3 = +/- pi/3 + 2pin?

I know this to be true but only from drawing out the unit circle and looking at it and the angles to equal pi/3. Is there some understanding I could gain to help me do this for other problems in the future?

Thanks