Question

6sinx*cosx-4cosx+sinx-2=0

Answer 1

$6sinx*cosx-4cosx+sinx-2=0 \frac{6sinx*cosx}{cosx} - \frac{4cosx}{cosx} + \frac{sinx}{cosx} - \frac{2}{cosx}$ 6sinx-4sinx-2=0 Sin x = t 6t-4t^2-2=0 D=b^2-4*a*c, D=16-4*6*(-2)=16+48=64. X1= 4+8/12=1, X2=4-8/12= - 1/3. Ответ: 1; -1/3.