Toán Lớp 11: 2sin2x – 3cos^2 x + 5sinxcosx -2 =0

1 bình luận về “Toán Lớp 11: 2sin2x – 3cos^2 x + 5sinxcosx -2 =0”

1. $$\begin{gathered} 2\sin 2x–3{\cos }^{2}x+5\sin x.\cos x–2=0 \\ <=>2\sin 2x–\frac{3}{2}–\frac{3}{2}\cos 2x+\frac{5}{2}\sin 2x–2=0 \\ <=>\frac{9}{2}\sin 2x–\frac{3}{2}\cos 2x–\frac{7}{2}=0 \\ <=>9\sin 2x–3\cos 2x=7 \\ <=>\frac{9}{3\sqrt{10}}\sin 2x–\frac{3}{3\sqrt{10}}\cos 2x=\frac{7}{3\sqrt{10}} \\ <=>\frac{3}{\sqrt{10}}\sin 2x–\frac{1}{\sqrt{10}}\cos 2x=\frac{7\sqrt{10}}{30} \\ <=>\sin 2x.\cos a–\cos 2x.\sin a=\frac{7\sqrt{10}}{30} \\ Với a là góc có \cos a=\frac{3}{\sqrt{10}}; \sin a=\frac{1}{\sqrt{10}} \\ <=>\sin \left(2x–a\right)=\frac{7\sqrt{10}}{30} \\ <=>2x–a=arc\sin \left(\frac{7\sqrt{10}}{30}\right)+k2\mathrm{\pi } \mathrm{hoặc} 2\mathrm{x}–\mathrm{a}=\mathrm{\pi }–\arcsin \left(\frac{7\sqrt{10}}{30}\right)+\mathrm{k}2\mathrm{\pi }, \mathrm{k}\in \mathrm{Z} \\ <=>\mathrm{x}=\frac{1}{2}\mathrm{a}+\frac{1}{2}\arcsin \left(\frac{7\sqrt{10}}{30}\right)+\mathrm{k\pi } \mathrm{hoặc} \mathrm{x}=\frac{\mathrm{\pi }}{2}+\frac{1}{2}\mathrm{a}+\frac{1}{2}+\arcsin \left(\frac{7\sqrt{10}}{30}\right)+\mathrm{k\pi }, \mathrm{k}\in \mathrm{Z} \\ \mathrm{Vậy} \mathrm{S}=\left\{\frac{1}{2}+\arcsin \left(\frac{7\sqrt{10}}{30}\right)+\mathrm{k\pi };\frac{\mathrm{\pi }}{2}+\frac{1}{2}\mathrm{a}+\frac{1}{2}+\arcsin \left(\frac{7\sqrt{10}}{30}\right)+\mathrm{k\pi }\right\}, \mathrm{k}\in \mathrm{Z} \end{gathered}$$

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