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

1 Answer
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