Решите тригонометрическое уравнение 1)2sin^2x+3sinx-5=0 2)10sin^2x-17cosx-16=0 3)5sin^2x+1sinxcosx+6cos^2x=0 4)3tgx-14ctgx+1=0

1) 2sin²x + 3sinx - 5 = 0 Пусть t = sinx, t ∈ [-1; 1]. 2t² + 3t - 5 = 0 D = 9 + 4•5•2 = 49 = 7² t1 = (-3 + 7)/4 = 4/4 = 1 t2 = (-3 - 7)/4 = -10/4 - не уд. условию Обратная замена: sinx = 1 x = π/2 + 2πn, n ∈ Z. 2) 10sin²x - 17cosx - 16 = 0 10 - 10cos²x - 17cosx - 16 = 0 -10cos²x - 17cosx - 6 = 0 10cos²x + 17cosx + 6 = 0 Пусть t = cosx, x ∈ [-1; 1]. D = 289 - 4•6•10 = 49 = 7² t1 = (-17 + 7)/20 = -10/20 = -1/2 t2 = (-17 - 7)/20 = -24/20 - не уд. условию Обратная замена: cosx = -1/2 x = ±arccos(-1/2) + 2πn, n ∈ Z x = ±2π/3 + 2πn, n ∈ Z. 3) 5sin²x + 17sinxcosx + 6cos²x = 0 Разделим на cos²x. 5tg²x + 17tgx + 6 = 0 Пусть t = tgx. D = 289 - 6•4•5 = 289 - 120 = 13² t1 = (-17 + 13)/10 = -4/10 = -2/5 t2 = (-17 - 13)/10 = -30/10 = -3 Обратная замена: tgx = -2/5 x = arctg(-2/5) + πn, n ∈ Z. x = arctg(-3) + πn, n ∈ Z. 4) 3tgx - 14ctg + 1 = 0 3tgx - 14/tgx + 1 = 0 3tg²x + tgx - 14 = 0 Пусть t = tgx. 3t² + t - 14 = 0 D = 1 + 14•4•3 = 13² t1 = (-1 + 13)/6 = 12/6 = 2 t2 = (-1 - 13)/6 = -14/6 = -7/3 обратная замена: tgx = 2 x = arctg2 + πn, n ∈ Z tgx = -7/3 x = arctg(-7/3) + πn, n ∈ Z.