Question

ребят, помогите)
2cos^2x+4cosx=3sin^2x

Answer 1

2cos^2(x)+4cos(x)=3sin^2(x)=3-3cos^2(x)
2cos^2(x)+4cos(x)=3-3cos^2(x)
5cos^2(x)+4cos(x)-3=0
cos(x)=t
5t^2+4t-3=0
t=cos(x)=(-2+корень(19))/5
t=cos(x)=(-2-корень(19))/5 - ложный корень
x = +/- arccos( (-2+корень(19))/5 ) + 2*pi*k