Question

Решите уравнение 2sin^2x+sinx-1=0

Answer 1

$2sin^2x+sinx-1=0\ sinx=t, t in [-1;1]\ 2t^2+t-1=0\ D=1-4*2(-1)=9\ t_1= frac{-1+3}{4}= frac{1}{2}\ t_2= frac{-1-3}{4}= -1\ sinx= frac{1}{2}\ x=(-1)^k arcsin frac{1}{2}+ pi k, k in Z \ x=(-1)^k frac{ pi }{6} + pi k, k in Z \ sinx=-1\ x=- frac{ pi }{2}+2 pi n, n in Z$