Question

a) 2cos(3x-п/2)=-1
б) sin(2x-п/4)=-√3/2

Answer 1

б) sin(2x-π/4)=√3/2
2x-π/4=(-1)^n *arcsin(√3/2)+πn, n∈Z
2x-π/4=(-1)^n *(π/3)+πn, n∈Z
2x=(-1)^n*(π/3)+π/4+πn |:2
x=(-1)^n *(π/6)+π/8+πn/2, n∈Z
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Сорян, не знаю а(