\(y=\left|sin^4x-2sin^2x+1+m\right|=\left|\left(sin^2x-1\right)^2+m\right|\)
Do \(0\le\left(sin^2x-1\right)^2\le1\)
TH1: \(m\ge0\Rightarrow y=\left(sin^2x-1\right)^2+m\ge m\Rightarrow m=2\)
TH2: \(-1\le m\le0\Rightarrow\left(sin^2x-1\right)^2+m=0\) có nghiệm
\(\Rightarrow\left|\left(sin^2x-1\right)^2+m\right|\ge0\) (có xảy ra dấu =) nên \(y_{min}=0\) ko thỏa mãn
TH3: \(m< -1\Rightarrow\left(sin^2x-1\right)^2+m< 0;\forall x\)
\(\Rightarrow y=\left|\left(sin^2x-1\right)^2+m\right|=-\left(sin^2x-1\right)^2-m\ge-1-m\)
\(\Rightarrow-1-m=2\Rightarrow m=-3\)
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