Ответ:
2sin2x−1=0sin2x=0.52x=(−1)k⋅6π+πk,k∈Zx=(−1)k⋅12π+2πk,k∈Z
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Ответ:
2sin2x−1=0sin2x=0.52x=(−1)k⋅6π+πk,k∈Zx=(−1)k⋅12π+2πk,k∈Z