Объяснение:
а) cos 2x = 2cos^2 x - 1, поэтому
8cos^4 x + 3(2cos^2 x - 1) - 6 = 0
8cos^4 x + 6cos^2 x - 3 - 6 = 0
8cos^4 x + 6cos^2 x - 9 = 0
Замена cos^2 x = y, заметим, что y ∈ [0; 1]
8y^2 + 6y - 9 = 0
D = 6^2 - 4*8(-9) = 36 + 288 = 324 = 18^2
y1 = (-6 - 18)/16 = -24/16 < 0 - не подходит
y2 = (-6 + 18)/16 = 12/16 = 3/4 ∈ [0; 1] - подходит
y = cos^2 x = 3/4
1) cos x = -√3/2;
x1 = 5П/6 + 2П*k; x2 = 7П/6 + 2П*k
2) cos x = √3/2;
x3 = П/6 + 2П*k; x4 = -П/6 + 2П*k
б) Промежутку [-7П/2; -2П] = [-21П/6; -12П/6] принадлежат корни:
x1 = 5П/6 - 4П = (5П - 24П)/6 = -19П/6
x2 = 7П/6 - 4П = (7П - 24П)/6 = -17П/6
x3 = -П/6 - 2П = (-П - 12П)/6 = -13П/6
1. 4y(2-y)-(y-5)²=8у-4у²-(у²-10у+25)= 8у-4у²-у²+10у-25=-5у²+18у-25
2. (c²+6)²-4(9-c)(1-3c)=с⁴+12с²+36-(12с²-112с+36)=с⁴+12с²+36-12с²+112с-36=с⁴+112с
3. (x-4)(x+4)-(x-3)²=х²-16-(х²-6х+9)=х²-16-х²+6х-9=6х-25