Question 891903
<pre>
{{{cos(3x)cos(5x)-cos(7x)cos(9x)}}}{{{""=""}}}{{{"0"}}}

We use the product to sum formula on each term

{{{cos(alpha)cos(beta)}}}{{{""=""}}}{{{expr(1/2)(cos(alpha+beta)+cos(alpha-beta))}}}

Swap the factors in each product so the difference will be positive:

{{{cos(5x)cos(3x)-cos(9x)cos(7x)}}}{{{""=""}}}{{{"0"}}} 

{{{expr(1/2)(cos(5x+3x)+cos(5x-3x))}}}{{{""-""}}}{{{expr(1/2)(cos(9x+7x)+cos(9x-7x))}}}{{{""=""}}}{{{"0"}}}

{{{expr(1/2)(cos(8x)+cos(2x))}}}{{{""-""}}}{{{expr(1/2)(cos(16x)+cos(2x))}}}{{{""=""}}}{{{"0"}}}

{{{(cos(8x)+cos(2x))-(cos(16x)+cos(2x))}}}{{{""=""}}}{{{"0"}}}

{{{cos(8x)+cos(2x)-cos(16x)-cos(2x)}}}{{{""=""}}}{{{"0"}}}

{{{cos(8x)+cross(cos(2x))-cos(16x)-cross(cos(2x))}}}{{{""=""}}}{{{"0"}}}

{{{cos(8x)-cos(16x)}}}{{{""=""}}}{{{"0"}}}

Use the doubla angle identity {{{cos(2theta)}}}{{{""=""}}}{{{2cos^2(theta)-1}}}
with {{{theta=8x}}}, {{{2theta=16x}}}

{{{cos(8x)-(2cos^2(8x)-1)}}}{{{""=""}}}{{{"0"}}}

{{{cos(8x)-2cos^2(8x)+1}}}{{{""=""}}}{{{"0"}}}

{{{-2cos^2(8x)+cos(8x)+1}}}{{{""=""}}}{{{"0"}}}

Multiply through by -1

{{{2cos^2(8x)-cos(8x)-1}}}{{{""=""}}}{{{"0"}}}

Factor the left side:

{{{(2cos(8x)+1)(cos(8x)-1)}}}{{{""=""}}}{{{"0"}}}

Use zero-factor property:

{{{2cos(8x)+1=0}}},     {{{cos(8x)-1=0}}}

{{{2cos(8x)=-1}}},     {{{cos(8x)=1}}}

{{{cos(8x)=-1/2}}},     {{{8x=2n*pi}}}
                    
{{{8x= pi +- pi/3+2n*pi}}},     {{{x=expr(n/4)*pi}}}       
 
{{{8x = (1 +- 1/3 + 2n)pi}}}

{{{8x= (3/3 +- 1/3 +6n/3)pi}}}

{{{8x = ((3 +- 1+6n)/3)pi}}}

{{{8x = ((6n +3 +- 1)/3)pi}}}

{{{x = ((6n +3 +- 1)/24)pi}}},     {{{x=expr(n/4)*pi}}}

Edwin</pre>