Question

Решие пж!Срочно надо.

Answer 1

$2sin^2x -3sin x + 1 = 0$      |   $t = sin x$
$2t^2 - 3t + 1 = 0$
$D = 9 - 8 = 1$
$t1 = 1 \\ t2 = \frac{1}{2}$
$sin x = 1 \\ sin x = \frac{1}{2}$
$x1 = \frac{Pi}{2} \\ x2 = \frac{Pi}{6} \\ x3 = \frac{5Pi}{6}$

$\frac{dy}{dx} (\frac{x}{x+1}) = \frac{1}{(x+1)^2}$
Значение в точке -2 равно:
$\frac{1}{(-2+1)^2} = 1$