Verified answer

б) sinx=y;

IyI≤1;

2y²-5y+2=0;

y=(5±√(25-16))/4=(5±3)/2? y=2>1, ∅; у=-2/4=-1/2;

sinx=-1/2; х=(-1)ⁿarcsin(-1/2)+πk; k∈Z; х=(-1)ⁿ⁺¹arcsin(1/2)+πk; k∈Z;

х=(-1)ⁿ⁺¹π/6+πk; k∈Z;

г) 4cos²x-4cosx+1=0

(2cosx-1)²=0;

cosx=y;

IyI≤1;

2y-1=0; 2y=1;

y=1/2;

cosx=1/2;

x=±arccos(1/2)+2πn; n∈Z

х=±π/3+2πn; n∈Z