Ответ: π/16+πn/2, π/24+πk/3, n,k∈Z

Объяснение:sin5x+cos5x=√2 cosx ║: √2;

1/√2·sin5x +1/√2· cos5x=cosx;

sinπ/4·sin5x+cos π/4·cos5x=cosx;

cos(5x - π/4)=cosx;

cos(5x -π/4)-cosx=0;

-2sin(5x-π/4-x)/2 ·sin (5x-π/4+x)/2 =0;

sin(2x-π/8)=0 ⇔2x-π/8=πn,n∈z ⇔x=π/16+πn/2,n∈z

                     sin(3x-π/8)=0 ⇔3x-π/8=πk,k∈z,  x=π/24+πk/3,k∈z

4. COS(arcsin1/4)

arcsin1/4=α,  sinα=1/4,α∈1 четверти

сos²α=1-sin²α,  cosα=+√(1-sin²α)=√(1-1/16)=√15/16=√15/4

cos(arcsin1/4)=√15/4