Sin2x+cosx = 0;
2sinx*cosx +cosx =0;
2cosx*(sinx +1/2) =0;
cosx=0 ⇒x=π/2 +2π*k , k∈Z;
sinx+1/2=0 ;
sinx =-1/2;
x= (-1)^(k+1) +π*k.
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(cos2x-1)tqx=0;
ОДЗ: x ≠ π/2 +π*k;
{cos2x-1=0 и tqx существует cos2x=1⇒2x =2π*k ⇔x=π*k;
tqx=0 ⇒ x=π*k