1) t = ± arccos a
1) a) 1 + cos x - 2sin²x = 0 Распишем 2sin²x = 2 - 2cos²x и подставим:
2cos²x + cos x -1 = 0
D = 1 + 8 = 9
cos x1 = (-1 + 3)\4 = 1\2 cos x2 = (-1 - 3)\4 = -1
x1 = ± π\3 + 2πk x2 = π + 2πk
б) sin2x + 2√3 cos²x = 0 Распишем sin2x = sin (x+x) = sinx cosx + cosx sinx = sinx (cosx + cosx) = 2sinx cosx и подставим:
2sinx cosx + 2√3 cos²x = 0
2cosx (sinx + √3 cosx) = 0
2cosx = 0 sinx + √3cosx = 0 \cosx
cosx = 0 tgx + √3 = 0
x = π\2 + πk tgx = - √3
x = - π\3 + πk