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дракон100
@дракон100
July 2022
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Решить уравнение:
3cos2x-sin2x+4sinx=0
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Kagoa
3cos²x-sin²x+4sinx=0cos²x=1-sin²x3-3sin²x-sin²x+4sinx=0-4sin²x+4sinx+3=0t=sinx -1≤t≤1-4t²+4t+3=0D=16+48=64 √D=8t₁=(-4+8)/(-8)= -(1/2)t₂=(-4-8)/(-8)=12/8 не подходит т.к. (12/8)>1sinx=-(1/2)x=(-1)ⁿ(-π/6)+nπ, n∈Z
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