Giải phương trình sau: |sinx + cosx | + sin2x = 1
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|sinx + cosx | + sin2x = 1 (1)
TH1:sinx + cosx ≤ 0
đặt t = sinx + cosx ( |t| ≤ 2√ 2)
=> sin2x = 2sinxcosx = t^2 - 1
pt(1) <=> |sinx + cosx | + sin2x = 1
<=> -(sinx + cosx ) + sin2x = 1
<=> 1 - t^2 + t = 1
<=> t^2 - t = 0
<=> t= 0 (t/m)
or t = 1 (loại)
=> sinx + cosx = 0
<=> √2 sin(x+π /4) = 0
<=> sin(x + π /4) = 0
<=> x + π /4 = π /2 + 2kπ
<=> x = π /4 + 2kπ
TH2 :sinx + cosx > 0
đặt t = sinx + cosx ( |t| ≤ 2√ 2)
=> sin2x = 2sinxcosx = t^2 - 1
pt(1) <=> |sinx + cosx | + sin2x = 1
<=> sinx + cosx + sin2x = 1
<=> t^2 -1 + t = 1
<=> t^2 + t - 2 = 0
=> t = 1
or t = -2 (loại )
=> sinx + cosx = 1
<=> √ 2sin(x+π /4) = 1
<=> sin(x + π /4) = 1/√ 2
<=> x + π/4 = π/4 + 2kπ
or x + π/4 = 3π/4 + 2kπ
<=> x = 2kπ
or x = π/2 + 2kπ