2sin^2 x - 10cos 2x = 9sin 2x +10 решите

2sin^2 x - 10cos 2x = 9sin 2x +102sin²x - 10(cos  ²x   - sin²x)  = 9*2sin xcosx +10*12sin²x - 10cos  ²x   +10 sin²x  = 9*2sin xcosx +10*(sin²x + cos²x) 2sin²x - 10cos  ²x   +10 sin²x  = 9*2sin xcosx +10sin ²x + 10cos²x2sin²x - 18sinxcosx -20cos²x = 0 sin²x - 9sinxcosx -10cos²x = 0 | : cos²x  ≠0 tg²x - 9tgx -10 = 0 по т. виета tgx = 10                                   tgx = -1 x = arctg10 +  πk , k  ∈ z           x = -π/4 +  πn , n  ∈z

2.d=6х в квадрате-4*1*4=36+16=52

хпервое=-6+52=48: 2=24

хвторое=-6-52=-58: 2=-29