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5 sin(2x-pi/6)-5=0

5 sin (2x -π/6) -5=05 (sin(2x -π/6) -1) =0sin(2x -π/6) -1=0sin(2x -π/6)=12x-π/6 =π/2 +2πn, n∈Z2x=π/6 +π/2 +2πn, n∈Z2x=4π/6 + 2πn, n∈Zx =4π/12 + πn, n∈Zx=π/3 +πn, n∈ZОтвет: π/3 +πn, n∈Z.