Решите тригонометрические уравнения: 1. 2cos^2x+3cosx-5=0 2. 6cos^2x-11sinx-10=0 3. Sin^2x+7sinx cosx+12cos^2x=0 4. 7tgx-8ctgx+10=0 5. 9cos^2x-sin^2x=8sinx cosx

1. 2cos²x + 3cosx - 5 = 0 Пусть t = cosx, t ∈ [-1; 1]. 2t² + 3t - 5 = 0 D = 9 + 2•4•5 = 49 = 7² t1 = (-3 + 7)/4 = 4/4 = 1 t2 = (-3 - 7)/4 = -10/4 = -2,5 - не уд. условию. Обратная замера: cosx = 1 x = 2πn, n ∈ Z 2. 6cos²x - 11sinx - 10 = 0 6 - 6sin²x - 11sinx - 10 = 0 -6sin²x - 11sinx - 4 = 0 6sin²x + 11sinx + 4 = 0 Пусть t = sinx, t ∈ Z. 6t² + 11t + 4 = 0 D = 121 - 4•6•4 = 25 = 5² t1 = (-11 + 5)/12 = -1/2 t2 = (-11 - 5)/12 = -16/12 - не уд. условию. Обратная замена: sinx = -1/2 x = (-1)ⁿ+¹π/6 + πn, n ∈ Z. 3. sin²x + 7sinxcosx + 12cos²x = 0 tg²x + 7tgx + 12 = 0 Пусть t = tgx. t² + 7t + 12 = 0 D = 49 - 48 = 1 t1 = (-7 + 1)/2 = -6/2 = -3 t2 = (-7 - 1)/2= -8/2 = -4 Обратная замена: tgx = -3 x = arctg(-3) + πn, n ∈ Z tgx = -4 x = arctg(-4) + πn, n ∈ Z. 4. 7tgx - 8ctgx + 10 = 0 7tgx - 8/tgx + 10 = 0 7tg²x + 10tgx - 8 = 0 (tgx ≠ 0) Пусть t = tgx. 7t² + 10t - 8 = 0 D = 100 + 4•7•8 = 324 = 18² t1 = (-10 + 18)/14 = 8/14 = 4/7 t2 = (-10 - 18)/14 = -28/14 = -2 Обратная замена: tgx = 4/7 x = arctg4/7 + πn, n ∈ Z tgx = -2 x = arctg(-2) + πn, n ∈ Z. 5. 9cos²x - sin²x = 8sinxcosx 9 - tg²x = 8tgx tg²x + 8tgx - 9 = 0 Пусть t = tgx. t² + 8t - 9 = 0 t1 + t2 = -8 t1•t2 = -9 t1 = -9 t2 = 1 Обратная замена: tgx = -9 x = arctg(-9) + πn, n ∈ Z. tgx = 1 x = π/4 + πn, n ∈Z.