Question #86ee3

2 Answers
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

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