Question

Question #86ee3

Answer 1

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`

Answer 2

`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`