A)4sin²x+√2tgx=0;⇒4sin²x+√2·sinx/cosx=0;cosx≠0;x≠π/2+kπ;k∈Z;⇒
4sin²x·cosx+√2·sinx=0;⇒2sin2x·sinx+√2·sinx=0;
√2·sinx·(√2sin2x+1)=0;
sinx=0;⇒x=kπ;k∈Z;
√2sin2x=-1;⇒sin2x=-1/√2;⇒
sin2x=-√2/2;⇒
2x=-π/4+2kπ;k∈Z;⇒
x=-π/8+kπ;k∈Z;
2x=5π/4+2kπ;k∈z;⇒
x=5π/8+kπ;k∈Z;
б)[-3π;-2π]⇒x=kπ;k∈Z;k=-3;k=-2;
x=-π/8+kπ;