Плиз можно без инетных решений​

sin(2x - π/3) = 1/√2

sin(2x - π/3) = √2/2

2x - π/3 = π/4 + 2πk  k ∈ Z

2x - π/3 = 3π/4 + 2πn n ∈ Z

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2x  = π/4 + π/3 + 2πk  k ∈ Z

2x  = 3π/4 + π/3 + 2πn n ∈ Z

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2x  = 7π/12 + 2πk  k ∈ Z

2x  = 13π/12  + 2πn n ∈ Z

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x  = 7π/24 + πk  k ∈ Z

x  = 13π/24  + πn n ∈ Z