Sin4x + корень из 3 * cos4x = корень из 2 ? делим лево право на 2 1/2sin4x+√3/2cos4x=√2/2 sin π/6sin4x+cosπ/6co4x=√2/2 cos(4x-π/6)=√2/2 4x-π/6=+-π/4+2πN 4x=-π/4+π/6+2πN 4x=-π/12+2πN x=-π/48+π/2N 4x=π/4+π/6+2πN 4x=5π/12+2πN x=5π/48+π/2N