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nếu câu trả lời hữu ích nhé!
`a)3sin4x+4cos4x=4`
`<=>3/5 sin4x + 4/5 cos 4x = 4/5`
Đặt `3/5=cos \alpha; 4/5 = sin \alpha` ta có:
`cos \alpha sin4x + sin \alpha cos 4x= sin \alpha`
`<=>sin(4x+\alpha)=sin\alpha`
`<=>` \(\left[ \begin{array}{l}4x+\alpha=\alpha+k2\pi\\4x+\alpha=\pi-\alpha+k2\pi\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}4x=k2\pi\\4x=\pi-2\alpha+k2\pi\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}-\dfrac{\alpha}{2}+\dfrac{k\pi}{2}\end{array} \right.\) `(k∈Z)`
`b,cos2x-2sin2x=1`
`<=>1/{\sqrt{5}} cos2x-2/{\sqrt{5}} sin2x=1/{\sqrt{5}}`
Đặt `1/{\sqrt{5}} = sin \alpha ; 2/{\sqrt{5}} = cos \alpha`
`<=>sin \alpha cos2x - cos \alpha sin 2x= sin \alpha`
`<=>sin(\alpha-2x)=sin\alpha`
`<=>` \(\left[ \begin{array}{l}\alpha-2x=\alpha+k2\pi\\\alpha-2x=\pi-\alpha+k2\pi\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}-2x=k2\pi\\x=\dfrac{-\pi}{2}+a-k\pi\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-k\pi\\x=\dfrac{-\pi}{2}+a-k\pi\end{array} \right.\) `(k∈Z)`
Hãy giúp mọi người biết câu trả lời này thế nào?
Bảng tin
6659
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oh
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gì v má =))
6659
451
6741
khôq kl -.-
2097
3702
2062
=))