How do you solve #4\cos x \cdot \cos 2x \cdot \cos 4x = \sin 8x#?

1 Answer
Ahmed A.
Apr 5, 2018

#x=30#

Explanation:

#4cos(x)*cos(2x)*cos(4x)=sin(8x)#

#sin2(A)=2sin(A) cos(A) # (Identity)

#sin(8x)=sin(2xx4x)# #(4x=A)#

#sin(2xx4x)=2cos(4x)sin(4x)#

#4cos(x)*cos(2x)*cos(4x)=2cos(4x)*sin(4x)#

Divide by #cos(4x)# both sides

#4cos(x)*cos(2x)=2sin(4x)#

Again

#sin(4x)=sin(2xx2x)# #(2x=A)#

#sin(2xx2x)=2cos(2x)sin(2x)#

#4cos(x)*cos(2x)=4cos(2x)*sin(2x)#

Divide by #cos(2x)# both sides

#4cos(x)=4sin(2x)# and divide by 4

#cos(x)=sin(2x)#

#sin(2x)=sin(2xxx)# #(x=A)#

#sin(2xxx)=2cos(x)*sin(x)#

#cos(x)=2cos(x)*sin(x)#

So, #sin(x)=1/2#

#x=30#

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