Question

8sin^2x+7sin2x+3cos2x+3=0 помогите решить

Answer 1

$8sin^{2}x+7sin(2x)+3cos(2x)+3=0$
$8sin^{2}x+7sin(2x)+3cos(2x)+3sin^{2}x+3cos^{2}x=0$
$11sin^{2}x+7*2sinx*cosx+3*(cos^{2}x-sin^{2}x)+3cos^{2}x=0$
$11sin^{2}x+14sinx*cosx+3cos^{2}x-3sin^{2}x+3cos^{2}x=0$
$8sin^{2}x+14sinx*cosx+6cos^{2}x=0$
$4sin^{2}x+7sinx*cosx+3cos^{2}x=0$
$4tg^{2}x+7tgx+3=0$

Замена: $tgx=t$
$4t^{2}+7t+3=0, D=49-4*4*3=1$
$t_{1}= \frac{-7+1}{8}=-\frac{3}{4}$
$t_{2}= \frac{-7-1}{8}=-1$

Вернемся к замене:
1) $tgx=-\frac{3}{4}$
$x=arctg(-\frac{3}{4})+ \pi k$
$x=-arctg(\frac{3}{4})+ \pi k$, k∈Z
2) $tgx=-1$
$x=-\frac{ \pi }{4}+ \pi k$, k∈Z

Answer 2

$8\sin^2x+7\sin2x+3\cos2x+3=0 \\ 8\sin^2x+14\sin x\cos x+3\cos^2x-3\sin^2x+3\cos^2x+3\sin^2x=0 \\ 8\sin^2x+14\sin x\cos x+6\cos^2x=0|:2\cos^2x \\ 4tg^2x+7tgx+3=0 \\ tg x=t \\ 4t^2+7t+3=0 \\ D=49-48=1 \\ t_1= \frac{-7+1}{8}= -0.75 \\ t_2=-1$

Возвращаемся к замене

$tg x=-1 \\ x_1=- \frac{\pi}{4} +\pi k,k \in Z \\ \\ tg x=-0.75 \\ x_2=-arctg0.75+ \pi k,k \in Z$