Toán Lớp 11: Giải phương trình ( sin2x + 1).( cân3 cosx – sinx -1)=0

1 bình luận về “Toán Lớp 11: Giải phương trình ( sin2x + 1).( cân3 cosx – sinx -1)=0”

1. $$\begin{gathered} \left(\sin 2x+1\right)\left(\sqrt{3}\cos x–\sin x–1\right)=0 \\ <=>\sin 2x+1=0 \left(1\right) hoặc \sqrt{3}\cos x–\sin x–1=0 \left(2\right) \\ \left(1\right)<=>\sin 2x=–1<=>2x=–\frac{\mathrm{\pi }}{2}+k2\mathrm{\pi }, \mathrm{k}\in \mathrm{Z} \\ <=>\mathrm{x}=–\frac{\mathrm{\pi }}{4}+\mathrm{k\pi }, \mathrm{k}\in \mathrm{Z} \\ \left(2\right)<=>\sqrt{3}\mathrm{cosx}–\mathrm{sinx}=1<=>\frac{\sqrt{3}}{2}\mathrm{cosx}–\frac{1}{2}\mathrm{sinx}=\frac{1}{2} \\ <=>\sin \frac{\mathrm{\pi }}{3}\mathrm{cosx}–\cos \frac{\mathrm{\pi }}{3}.\mathrm{sinx}=\frac{1}{2} \\ <=>\sin \left(\frac{\mathrm{\pi }}{3}–\mathrm{x}\right)=\frac{1}{2} \\ <=>\frac{\mathrm{\pi }}{3}–\mathrm{x}=\frac{\mathrm{\pi }}{6}+\mathrm{k}2\mathrm{\pi } \mathrm{hoặc} \frac{\mathrm{\pi }}{3}–\mathrm{x}=\mathrm{\pi }–\frac{\mathrm{\pi }}{6}+\mathrm{k}2\mathrm{\pi }, \mathrm{k}\in \mathrm{Z} \\ <=>\mathrm{x}=\frac{\mathrm{\pi }}{6}+\mathrm{k}2\mathrm{\pi } \mathrm{hoặc} \mathrm{x}=–\frac{\mathrm{\pi }}{2}+\mathrm{k}2\mathrm{\pi }, \mathrm{k}\in \mathrm{Z} \\ \mathrm{vậy} \mathrm{S}=\left\{–\frac{\mathrm{\pi }}{4}+\mathrm{k\pi };\frac{\mathrm{\pi }}{6}+\mathrm{k}2\mathrm{\pi };–\frac{\mathrm{\pi }}{2}+\mathrm{k}2\mathrm{\pi }\right\}, \mathrm{k}\in \mathrm{Z} \end{gathered}$$

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