Решите уравнение sin3x+cos2x=1

Sin3x+cos2x=1

3sinx-4sin³x+1-2sin²x=1

sinx=t€[-1; 1]

4t³+2t²-3t=0

t (4t²+2t-3) = 0

1) t=0; sinx=0; x=πn

2) 4t²+2t-3=0

D=4+48=52

t = (-2±√52) / 8=2 (-1±√13) / 8

= (-1±√13) / 4

t1<-1

t2 = (-1+√13) / 4

sinx = (-1+√13) / 4

x = (-1) ⁿarcsin (-1+√13) / 4+πn;

n€Z