Question

1) 2cos^2-cosx-1=0
2) 2sinxcosx=cos2x-2sin^2x
3) 1-cos2x= sin2x
Помогите решить очень срочно!

Answer 1

$2cos^2x - cosx - 1 = 0$
Пусть $t = cosx, t in [-1; 1]$
$2t^2 - t - 1 = 0 \ D = 1 + 2 cdot 4= 9 = 3^2 \ t_1 = frac{1 + 3}{4} = 1 \ t_2= frac{1 - 3}{4} = - frac{1}{2}$
Обратная замена:
$cosx = 1 \ x = 2 pi n, n in Z\ \ cosx = - frac{1}{2} \ x = pm frac{2 pi }{3} + 2 pi n, n in Z$

$2sinxcosx = cos2x - 2sin^2x \ 2sinxcosx = cos^2x - sin^2x - 2sin^2x \ 3sin^2x + 2sinxcosx - cos^2x = 0 \ 3tg^2x + 2tgx - 1 = 0$
Пусть $t = tgx.$
$3t^2 + 2t - 1 = 0 \ D = 4 + 3 cdot 4 = 16 = 4^2 \ t_1 = frac{-2 + 4}{6} = frac{1}{3} \ \ t_2 = frac{-2 - 4}{6} = -1$
Обратная замена:
$tgx = frac{1}{3} \ x = arctg frac{1}{3} + pi n, n in Z \ \ tgx = -1 \ x = - frac{ pi }{4} + pi n, n in Z$

$1 - cos2x = sin2x \ 1 - (1 - 2sin^2x) = 2sinxcosx \ 2sin^2x - 2sinxcosx = 0 \ sin^2x - sinxcosx = 0 \ sinx(sinx - cosx) = 0 \ sinx = 0 sinx - cosx = 0 \ x = pi n, in Z tgx - 1 = 0 \ . x = frac{ pi }{4} + pi n, n in Z$