Question

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

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:
$tgx = \frac{1}{3} \\ x = arctg \frac{1}{3} + \pi n, n \in Z \\ \\ tgx = -1 \\ x = - \frac{ \pi }{4} + \pi n, n \in Z$

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