sin(2x)* sin(6x) = cos(x)*cos(3x) ;
( сos(6x-2x) - cos(6x+2x) ) / <em>2 </em>= ( cos(3x+x) +cos(3x-x) ) / 2 ;
<em>cos4x </em>-cos8x = cos<em>4x</em> + cos2x<em></em>;
cos8x +cos2x =0 ;
2cos( (8x+2x)/2) *cos( (8x-2x)/2)=0 ;
cos5x*cos3x =0 ;
cos5x = 0 ⇒5x =π/2 +πk , k ∈ℤ ⇔x =π/10 +πk/5 ,k∈ℤ
или
cos3x =0 ⇒3x =π/2 +πk , k ∈ℤ ⇔x =π/6 +πk/3 ,k∈ℤ