If \( 1+\cos2x+\cos4x+\cos6x=k\cos x\cos2x\cos3x \), then \( k= \)
Solution:
\[
1+\cos2x
=
2\cos^2x
\]
Therefore,
\[
1+\cos2x+\cos4x+\cos6x
\]
\[
=
2\cos^2x+(\cos4x+\cos6x)
\]
Using identity,
\[
\cos A+\cos B
=
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\]
\[
=
2\cos^2x+2\cos5x\cos x
\]
\[
=
2\cos x(\cos x+\cos5x)
\]
Again using,
\[
\cos A+\cos B
=
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\]
\[
=
2\cos x(2\cos3x\cos2x)
\]
\[
=
4\cos x\cos2x\cos3x
\]
Comparing with
\[
k\cos x\cos2x\cos3x
\]
\[
k=4
\]
\[
\boxed{4}
\]