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#
