Question #86ee3

2 Answers
Douglas K.
Apr 11, 2017

Use the identity #cos^2(A)-sin^2(A) = cos(2A)#
Use the inverse cosine on both sides
Solve for x with repeating values.

Explanation:

Given: #cos^2(2x)-sin^2(2x)=sqrt3/2#

Use the identity #cos^2(A)-sin^2(A) = cos(2A)#

#cos(4x) = sqrt(3)/2#

Use the inverse cosine on both sides:

#4x = cos^-1(sqrt(3)/2)#

We know the values for this inverse cosine:

#4x = pi/6 and 4x = (11pi)/6#

We know that the two repeat every #2npi#

#4x = pi/6 + 2npi and 4x = (11pi)/6 + 2npi; n in ZZ#

Divide both equations by 4:

#x = pi/24 + npi/2 and x = (11pi)/24 + npi/2; n in ZZ#

Nghi N
Apr 11, 2017

#pi/12 + kpi#
#(11pi)/12 + kpi#

Explanation:

Use trig identity:
#cos^2 x - sin^2 x = cos 2x#
In this case:
#cos^2 x - sin^2 x = cos 2x = sqrt3/2#
Trig table and unit circle give 2 solutions:
#2x = +- pi/6 + 2kpi#

a. #2x = pi/6 + 2kpi# --> #x = pi/12 + kpi#
b. #2x = - pi/6 + 2kpi#, or #2x = (11pi)/6 + 2kpi# (co-terminal)
x = #(11pi)/12 + kp#

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