1.Nghiệm phương trình sinx + 4cosx = 2 + sin2x là? 2.Phương trình √2 (sinx- 2cosx) = 2 -sin2x có hai họ nghiệm có dạng x = α + k2π; x = β + k2π ( 0 < hoặc=α,β

youngheunyaaa

2 năm trước

1.Nghiệm phương trình sinx + 4cosx = 2 + sin2x là? 2.Phương trình √2 (sinx- 2cosx) = 2 -sin2x có hai họ nghiệm có dạng x = α + k2π; x = β + k2π ( 0 < hoặc=α,β < hoặc =π ) .Khi đó α.β Bằng?

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ahihi

Đáp án + Giải thích các bước giải:

`1.`

`sinx + 4cosx = 2 + sin2x`

`<=>sinx+4cosx-2-2sinxcosx=0`

`<=>(sinx-2sinxcosx)-2+4cosx=0`

`<=>sinx(1-2cosx)-2(1-2cosx)=0`

`<=>(sinx-2)(1-2cosx)=0`

`<=>` $\left[ \begin{array}{l}\sin x=2(VN)\\\cos x=\dfrac{1}{2}\end{array} \right.$

`<=>x=+-pi/3+k2pi(kinZZ)`

`2.`

`sqrt2 (sinx- 2cosx) = 2 -sin2x`

`<=>sqrt2sinx-2sqrt2cosx+2sinxcosx-2=0`

`<=>sqrt2sinx(1+sqrt2cosx)-2(1+sqrt2cosx)=0`

`<=>(1+sqrt2cosx)(sqrt2sinx-2)=0`

`<=>`$\left[ \begin{array}{l}\cos x=-\dfrac{1}{\sqrt2}\\\sin x=\sqrt2(VN)\end{array} \right.$

`<=>x=+-(3pi)/4+k2pi(kinZZ)`

`=>alpha=(3pi)/4,beta=-(3pi)/4`

`=>alpha.beta=-(9pi^2)/16`

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nngocanh0610

Đáp án:

$1) x=\pm \dfrac{\pi}{3} +k 2 \pi(k \in \mathbb{Z})\\ 2)\alpha.\beta=-\dfrac{9 \pi^2}{16}$

Giải thích các bước giải:

$1)\sin x+4\cos x=2+\sin 2x\\ \Leftrightarrow \sin x+4\cos x-2-\sin 2x=0\\ \Leftrightarrow \sin x+4\cos x-2-2\sin x\cos x=0\\ \Leftrightarrow \sin x-2-2\sin x\cos x+4\cos x=0\\ \Leftrightarrow \sin x-2-2\cos x(\sin x-2)=0\\ \Leftrightarrow (1-2\cos x)(\sin x-2)=0\\ \Leftrightarrow \left[\begin{array}{l} \cos x=\dfrac{1}{2}\\ \sin x=2(\text{Vô nghiệm})\end{array} \right.\\ \Leftrightarrow \cos x=\dfrac{1}{2}\\ \Leftrightarrow \cos x=\cos\left(\dfrac{\pi}{3}\right)\\ \Leftrightarrow x=\pm \dfrac{\pi}{3} +k 2 \pi(k \in \mathbb{Z})\\ 2)\\ \sqrt{2}(\sin x-2\cos x)=2-\sin 2x\\ \Leftrightarrow \sqrt{2}\sin x-2\sqrt{2}\cos x-2+\sin 2x=0\\ \Leftrightarrow \sqrt{2}\sin x-2\sqrt{2}\cos x-2+2\sin x\cos x=0\\ \Leftrightarrow \sqrt{2}\sin x-2+2\sin x\cos x-2\sqrt{2}\cos x=0\\ \Leftrightarrow \sqrt{2}(\sin x- \sqrt{2})+2\cos x(\sin x -\sqrt{2})=0\\ \Leftrightarrow (\sqrt{2}+2\cos x)(\sin x -\sqrt{2})=0\\ \Leftrightarrow \left[\begin{array}{l}\cos x=-\dfrac{\sqrt{2}}{2}\\ \sin x =\sqrt{2}(\text{Vô nghiệm})\end{array} \right.\\ \Leftrightarrow \cos x=-\dfrac{\sqrt{2}}{2}\\ \Leftrightarrow \cos x=\cos\left(\dfrac{3\pi}{4}\right)\\ \Leftrightarrow \left[\begin{array}{l} x=\dfrac{3\pi}{4}+k 2 \pi(k \in \mathbb{Z})\\ x=-\dfrac{3\pi}{4}+k 2 \pi(k \in \mathbb{Z})\end{array} \right.\\\Rightarrow \alpha.\beta=-\dfrac{9 \pi^2}{16}$

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