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=cosxcos3x

2*sinxcosx*2sin3xcos3x-cosxcos3x=0
cosxcos3x(4sinxsin3x-1)=0

1)cosx=0;x=π/2+πn
2)cos3x=0;3x=π/2+πk;x=π/6+πk/3

3)4sinxsin3x=1
sinxsin3x=1/4
sinx*(3sinx-4sin³x)=1/4
3sin²x-4sin⁴x-1/4=0
16sin⁴x-12sin²x+1=0
sin²x=t>0
16t²-12t+1=0
D/4=36-16=20
t=(6±2√5)/16=(3±√5)/8
t1=(3-√5)/8
t2=(3+√5)/8
sin²x=(3+√5)/8
sinx=√(3-√5)/8=√(6-2√5)/16=
√(√5)²-2√5+1)/16=(√5-1)²/16=(√5-1)/4
sinx=(√5-1)/4
x=(-1)ⁿarcsin((√5-1)/4)+πn

sinx=(√5+1)/4
x=(-1)ⁿarcsin((√5+1)/4)+πn