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

1) 2sin²x + 3sinx - 5 = 0Пусть t = sinx, t ∈ [-1; 1].2t² + 3t - 5 = 0D = 9 + 4•5•2 = 49 = 7²t1 = (-3 + 7)/4 = 4/4 = 1t2 = (-3 - 7)/4 = -10/4 - не уд. условиюОбратная замена:sinx = 1x = π/2 + 2πn, n ∈ Z.2) 10sin²x - 17cosx - 16 = 010 - 10cos²x - 17cosx - 16 = 0-10cos²x - 17cosx - 6 = 010cos²x + 17cosx + 6 = 0Пусть t = cosx, x ∈ [-1; 1].D = 289 - 4•6•10 = 49 = 7²t1 = (-17 + 7)/20 = -10/20 = -1/2t2 = (-17 - 7)/20 = -24/20 - не уд. условиюОбратная замена:cosx = -1/2x = ±arccos(-1/2) + 2πn, n ∈ Zx = ±2π/3 + 2πn, n ∈ Z.3) 5sin²x + 17sinxcosx + 6cos²x = 0Разделим на cos²x.5tg²x + 17tgx + 6 = 0Пусть t = tgx.D = 289 - 6•4•5 = 289 - 120 = 13²t1 = (-17 + 13)/10 = -4/10 = -2/5t2 = (-17 - 13)/10 = -30/10 = -3Обратная замена:tgx = -2/5x = arctg(-2/5) + πn, n ∈ Z.x = arctg(-3) + πn, n ∈ Z.4) 3tgx - 14ctg + 1 = 03tgx - 14/tgx + 1 = 03tg²x + tgx - 14 = 0Пусть t = tgx.3t² + t - 14 = 0D = 1 + 14•4•3 = 13²t1 = (-1 + 13)/6 = 12/6 = 2t2 = (-1 - 13)/6 = -14/6 = -7/3обратная замена:tgx = 2x = arctg2 + πn, n ∈ Ztgx = -7/3x = arctg(-7/3) + πn, n ∈ Z.