Question

Помогите решить 4cos^2 x-3sin x=3

Answer 1

$4cos^2x-3sin x=3\4(1-sin^2x)-3sin x=3\4-4sin^2x-3sin x=3\4sin^2x+3sin x-1=0\sin x=t,;sin^2x=t^2,;-1leq tleq1\4t^2+3t-1=0\D=9+4cdot4=25\t_{1,2}=frac{-3pm5}8\t_1=-1,;t_2=frac14\sin x=-1Rightarrow x=-fracpi2+2pi n\sin x=frac14Rightarrow x=(-1)^ncdotarcsinleft(frac14right)+pi n,;ninmathbb{Z}$