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НАЙДИТЕ х : sin(2x) sin(6x) = cos(x)cos(3x)

sin(2x)* sin(6x) = cos(x)*cos(3x) ;

( сos(6x-2x) - cos(6x+2x) ) / 2 = ( cos(3x+x) +cos(3x-x) ) / 2 ;

cos4x -cos8x = cos4x + cos2x ;

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

sin2xsin6x=cosxcos3x2*sinxcosx*2sin3xcos3x-cosxcos3x=0cosxcos3x(4sinxsin3x-1)=01)cosx=0;x=π/2+πn2)cos3x=0;3x=π/2+πk;x=π/6+πk/33)4sinxsin3x=1sinxsin3x=1/4sinx*(3sinx-4sin³x)=1/43sin²x-4sin⁴x-1/4=016sin⁴x-12sin²x+1=0sin²x=t>016t²-12t+1=0D/4=36-16=20t=(6±2√5)/16=(3±√5)/8t1=(3-√5)/8t2=(3+√5)/8sin²x=(3+√5)/8sinx=√(3-√5)/8=√(6-2√5)/16=√(√5)²-2√5+1)/16=(√5-1)²/16=(√5-1)/4sinx=(√5-1)/4x=(-1)ⁿarcsin((√5-1)/4)+πnsinx=(√5+1)/4x=(-1)ⁿarcsin((√5+1)/4)+πn