решить
4sin² x + 13sin x cos x + 10cos2 x = 0
2(1-cos2x) +( 13sin2 x)/2 + 10cos2 x = 0
4-4cos2x+13sin2 x+10cos2 x = 0
6cos2x+13sin2 x=-4
[6/(√(6²+13²) ]cos2x+[13/(√(6²+13²) ]sin2 x=-4/(√(6²+13²)
(6/√215)cos2x+[13/(√(215) ]sin2 x=-4/(√(215)
cos(2x-φ)=-4/(√(215), φ=arctg(13/6)
1) 2x-φ=π-arccos(4/√215)+πn, n∈Z
x=[π-arccos(4/√215)+φ+πn]/2, n∈Z
2) 2x-φ=-(π-arccos(4/√215))+πn, n∈Z x=[-π+arccos(4/√215)+φ+πn]/2, n∈Z