Question

помогите решить:

1)2cos2x=sinx-1

Answer 1

$2cos2x=sinx-1\2(1-2sin^2x)=sinx-1\2-4sin^2x-sinx+1=0\-4sin^2x-sinx+3=0\4sin^2x+sinx-3=0\sinx=t;-1leq t leq1\4t^2+t-3=0\sqrt{D}=sqrt{1+48}=7\x_{1}=frac{-1+7}{8}=frac{6}{8}=frac{3}{4};x_{2}=frac{-1-7}{8}=-1\sinx=-1 sinx=frac{3}{4}\x=-frac{pi}{2}+2pi*n x=(-1)^karcsinfrac{3}{4}+pi*k$