# How do you solve cos [2(x-pi/3)]=sqrt3/ 2?

May 7, 2016

x = π/2

#### Explanation:

Given

cos(2(x- (π/3)))= ((3)^(1/2)/2)

let Ω =  (2(x- (π/3)),

then

cos Ω = ((3)^(1/2)/2)

we notice that $\left({\left(3\right)}^{\frac{1}{2}} / 2\right)$ is a special angle and hence,

the angle Ω that gives cos Ω = ((3)^(1/2)/2) must be π/3

Therefore, Ω =π/3, equate Ω to solve for x,

 (2(x- (π/3)) = π/3

x =(π/3)/2+(π/3)

= π/6 + π/3

=π/2

May 7, 2016

$x = \frac{5 \pi}{12} + 2 k \pi$
$x = \frac{\pi}{4} + 2 k \pi$

#### Explanation:

$\cos \left(2 x - \frac{2 \pi}{3}\right) = \frac{\sqrt{3}}{2}$
Trig table and unit circle -->
$\left(2 x - \frac{2 \pi}{3}\right) = \pm \frac{\pi}{6}$
a. $2 x - \frac{2 \pi}{3} = \frac{\pi}{6} - \to 2 x = \frac{\pi}{6} + \frac{2 \pi}{3} = \frac{5 \pi}{6} \to x = \frac{5 \pi}{12}$
b. $2 x - \frac{2 \pi}{3} = - \frac{\pi}{6} - \to 2 x = \frac{2 \pi}{3} - \frac{\pi}{6} = \frac{3 \pi}{6} = \frac{\pi}{2}$ -->
$x = \frac{\pi}{4}$
$x = \frac{5 \pi}{12} + 2 k \pi$
$x = \frac{\pi}{4} + 2 k \pi$
$x = \frac{5 \pi}{12}$ --> cos (2x - (2pi)/3) = cos (6pi/12 - (4pi)/12 =
$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$. OK
$x = \frac{\pi}{4}$ --> $\cos \left(2 x - \frac{2 \pi}{3}\right) = \cos \left(\frac{\pi}{2} - \frac{2 \pi}{3}\right) =$
$= \cos \left(- \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$. OK