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

2 Answers
Peter
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 #((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#

Nghi N.
May 7, 2016

#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

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