Toán Lớp 11: Giải pt 2(cosx + √3 sinx) cosx = cosx – √3 sinx +1

1 bình luận về “Toán Lớp 11: Giải pt 2(cosx + √3 sinx) cosx = cosx – √3 sinx +1”

1. $$\begin{gathered} 2\left(\cos x+\sqrt{3}\sin x\right).\cos x=\cos x–\sqrt{3}\sin x+1 \\ <=>2{\cos }^{2}x–1+2\sqrt{3}\sin x.\cos x=\cos x–\sqrt{3}\sin x \\ <=>\cos 2x+\sqrt{3}\sin 2x=\cos x–\sqrt{3}\sin x \\ <=>\frac{1}{2}\cos 2x+\frac{\sqrt{3}}{2}\sin 2x=\frac{1}{2}\cos x–\frac{\sqrt{3}}{2}\sin x \\ <=>\cos 2x.\cos \frac{\mathrm{\pi }}{3}+\sin 2x.\sin \frac{\mathrm{\pi }}{3}=\cos x.\cos \frac{\mathrm{\pi }}{3}–\sin x.\sin \frac{\mathrm{\pi }}{3} \\ <=>\cos \left(2x–\frac{\mathrm{\pi }}{3}\right)=\cos \left(x+\frac{\mathrm{\pi }}{3}\right) \\ <=>2x–\frac{\mathrm{\pi }}{3}=x+\frac{\mathrm{\pi }}{3}+k2\mathrm{\pi } hoặc 2x–\frac{\mathrm{\pi }}{3}=–x–\frac{\mathrm{\pi }}{3}+k2\mathrm{\pi }, \mathrm{k}\in \mathrm{Z} \\ <=>\mathrm{x}=\frac{2\mathrm{\pi }}{3}+\mathrm{k}2\mathrm{\pi } \mathrm{hoặc} 3\mathrm{x}=\mathrm{k}2\mathrm{\pi }, \mathrm{k}\in \mathrm{Z} \\ <=>\mathrm{x}=\frac{2\mathrm{\pi }}{3}+\mathrm{k}2\mathrm{\pi } \mathrm{hoặc} \mathrm{x}=\frac{\mathrm{k}2\mathrm{\pi }}{3}, \mathrm{k}\in \mathrm{Z} \\ <=>\mathrm{x}=\frac{\mathrm{k}2\mathrm{\pi }}{3}, \mathrm{k}\in \mathrm{Z} \\ \mathrm{Vậy} \mathrm{x}=\frac{\mathbf{k}\mathbf{2}\mathbf{\pi }}{\mathbf{3}}\mathbf{,}\mathbf{ }\mathbf{k}\mathbf{\in }\mathbf{Z} \end{gathered}$$

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