Sin (2|arcsinx|)=1

sin (2|arcsinx|)=12|arcsinx|=π/2+2πk|arcsinx|=π/4+πkarcsinx=π/4+πk при arcsinx≥0 arcsinx=-π/4-πk при arcsinx<0 то есть х∈[0; 1] то есть х∈[-1; 0) х=sin (π/4+πk) х=sin (-π/4-πk) x1=sinπ/4=√2/2∈[0; 1] x1=sin (-π/4)=-√2/2∈[-1; 0) x2=sin5π/4=-√2/2∉[0; 1] x2=sin3π/4=-√2/2∉[-1; 0) x3=sin (-3π/4)=-√2/2∉[0; 1] x3=sin (-5π/4)=√2/2∈[-1; 0) ответ x=±√2/2