For this problem, we will be using the following:
- a^2-b^2 = (a+b)(a-b)
- cos^2(theta)+sin^2(theta) = 1
- cos^2(theta) - sin^2(theta) = cos(2theta)
- sin(theta) = cos(theta - pi/2)
- cos(theta) - cos(gamma) = -2sin((theta-gamma)/2)sin((theta+gamma)/2)
- ab = 0 <=> a = 0 or b = 0
- sin(theta) = 0 <=> theta = npi, n in ZZ
Note that aside from some basic arithmetic, the following steps use the above in the order they are listed. Proceeding:
cos^4(2x) = sin(3x) + sin^4(2x)
<=> cos^4(2x) - sin^4(2x) = sin(3x)
<=> (cos^2(2x)+sin^2(2x))(cos^2(2x)-sin^2(2x))=sin(3x)
<=>(1)(cos^2(2x)-sin^2(2x))=sin(3x)
<=>cos(2(2x)) = sin(3x)
<=> cos(4x) = sin(3x)
<=> cos(4x) = cos(3x-pi/2)
<=> cos(4x)-cos(3x-pi/2) = 0
<=> -2sin((7x-pi/2)/2)sin((x+pi/2)/2)=0
<=>-2sin(7/2x-pi/4)sin(x/2+pi/4)=0
<=>sin(7/2x-pi/4) = 0 or sin(x/2+pi/4) = 0
Case 1: sin(7/2x-pi/4) = 0
=> 7/2x-pi/4 = npi, n in ZZ
=> 7/2x = (n+1/4)pi, n in ZZ
=> x = (2pi)/7(n+1/4), n in ZZ
Case 2: sin(x/2+pi/4) = 0
=> x/2+pi/4 = npi, n in ZZ
=> x/2 = (n-1/4)pi, n in ZZ
=> x = 2pi(n-1/4), n in ZZ
Putting the two together, we get our final result:
x in {(2pi)/7(n+1/4), 2pi(n-1/4)} where n in ZZ
