ответ: x =π/6±π/4+2πk; k∈Z
Объяснение:
2cos(x -π/6)-√2=0
cos(x -π/6)=√2/2
(x -π/6)=±arccos(√2/2)+2πk; k∈Z
(x -π/6)=±π/4+2πk; k∈Z
x =π/6±π/4+2πk; k∈Z
ответ: x =π/6±π/4+2πk; k∈Z
Объяснение:
2cos(x -π/6)-√2=0
cos(x -π/6)=√2/2
(x -π/6)=±arccos(√2/2)+2πk; k∈Z
(x -π/6)=±π/4+2πk; k∈Z
x =π/6±π/4+2πk; k∈Z