Question #86ee3

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Question #86ee3

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Answer 1 · 2017-04-11T02:43:11

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 · 2017-04-11T13:37:10

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

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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Question #86ee3

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