Bài2:\(Pt\Leftrightarrow\left(3m-2\right)\left(1-\cos2x\right)-\left(10m-4\right)\sin2x+3\left(2m+1\right)\left(1+\cos2x\right)=0.\)\(\Leftrightarrow\cos2x\left(3m+5\right)-\sin2x\left(10m-4\right)+9m+1=0\)Pt vô nghệm \(\Leftrightarrow\left(10m-4\right)^2+\left(3m+5\right)^2-\left(9m+1\right)^2< 0\Leftrightarrow28m^2-68m+40< 0\)\(\Leftrightarrow\frac{10}{7}>m>1\)Bài 3:\(pt\Leftrightarrow\cos2x.\cos x+\cos2x-m.\cos^2x-m.\cos x-m.\sin^2x=0\)\(\Leftrightarrow\cos x.\cos2x+\cos2x-m.\cos...
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Bài2:
\(Pt\Leftrightarrow\left(3m-2\right)\left(1-\cos2x\right)-\left(10m-4\right)\sin2x+3\left(2m+1\right)\left(1+\cos2x\right)=0.\)
\(\Leftrightarrow\cos2x\left(3m+5\right)-\sin2x\left(10m-4\right)+9m+1=0\)
Pt vô nghệm \(\Leftrightarrow\left(10m-4\right)^2+\left(3m+5\right)^2-\left(9m+1\right)^2< 0\Leftrightarrow28m^2-68m+40< 0\)
\(\Leftrightarrow\frac{10}{7}>m>1\)
Bài 3:
\(pt\Leftrightarrow\cos2x.\cos x+\cos2x-m.\cos^2x-m.\cos x-m.\sin^2x=0\)
\(\Leftrightarrow\cos x.\cos2x+\cos2x-m.\cos x-m=0\Leftrightarrow\left(\cos x+1\right)\left(\cos2x-m\right)=0\Leftrightarrow\orbr{\begin{cases}\cos x=-1\Leftrightarrow x=\left(2k+1\right)\pi\\\cos2x=m\end{cases}}\) Th1: \(x=\left(2k+1\right)\pi\Rightarrow0\le\left(2k+1\right)\pi\le\frac{2\pi}{3}\Leftrightarrow-\frac{1}{2}\le k\le-\frac{1}{6}\Leftrightarrow\) không có k thỏa mãn
=> TH1, pt ko có ngh trên khoảng đã cho
Th2: \(\cos2x=m\) \(x\in\left[0,\frac{\pi}{3}\right]\Rightarrow2x\in\left[0,\frac{2\pi}{3}\right]\)
BBT của hs y=cos2x trên khoảng đã cho:
x 2x y
=> phương trinh có 2 ngh trên khoảng [0,2pi/3] \(\Leftrightarrow-1< m\le-\frac{1}{2}\)
