Question

помогите решить пример

Answer 1

$frac{1}{cos(frac{3}2pi-x)}-frac{1}{sin^2x}=-2\ frac{1}{-sinx}-frac{1}{sin^2x}=-2\ t=frac{1}{sinx}\ -t-t^2=-2\ t^2+t-2=0\ t_1=1; t_2=-2\ begin{cases} frac{1}{sinx}=1\ frac{1}{sinx}=-2 end{cases}\ \ begin{cases} sinx=1; sinxneq0\ sinx=-frac{1}2; end{cases}\ \ begin{cases} x=pipmpi n\ x=-frac{pi}{6}pm2pi n; x=-frac{5}6pipm2pi n\ end{cases}$
$xin[pi;frac{5}2pi] = textgreater x_1=pi; x_2=1frac{1}6pi; x_3=1frac{5}6pi; x_4=frac{5}2pi$