Question

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

Answer 1

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 `((3)^(1/2)/2)` 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`

Answer 2

`x = (5pi)/12 + 2kpi`
`x = pi/4 + 2kpi`

Explanation:

`cos (2x - (2pi)/3) = sqrt3/2`
Trig table and unit circle -->
`(2x - (2pi)/3) = +- pi/6`
a. `2x - (2pi)/3 = pi/6 --> 2x = pi/6 + (2pi)/3 = (5pi)/6 -> x = (5pi)/12`
b. `2x - (2pi)/3 = -pi/6 --> 2x = (2pi)/3 - pi/6 = (3pi)/6 = pi/2` -->
`x = pi/4`
General answers:
`x = (5pi)/12 + 2kpi`
`x = pi/4 + 2kpi`
Check.
`x = (5pi)/12` --> `cos (2x - (2pi)/3) = cos (6pi/12 - (4pi)/12 =`
`cos (pi/6) = sqrt3/2`. OK
`x = pi/4` --> `cos (2x - (2pi)/3) = cos (pi/2 - (2pi)/3) =`
`= cos (-pi/6) = sqrt3/2`. OK