Toán Lớp 11: 6sin²2x+sin2x-1=0 Sin⁴x-cos⁴x=1/2

1 bình luận về “Toán Lớp 11: 6sin²2x+sin2x-1=0 Sin⁴x-cos⁴x=1/2”

1. $$\begin{gathered} 6{\sin }^{2}2x+\sin 2x–1=0 \\ Đặt t=\sin 2x, t\in \left[–1;1\right] \\ Ta có: 6t^2+t–1=0 \\ Giải ra được t=–\frac{1}{2} \left(tm\right) hoặc t=\frac{1}{3} \left(tm\right) \\ t=–\frac{1}{2}<=>\sin 2x=–\frac{1}{2} \\ <=>2x=–\frac{\mathrm{\pi }}{6}+k2\mathrm{\pi } \mathrm{hoặc} 2\mathrm{x}=\mathrm{\pi }+\frac{\mathrm{\pi }}{6}+\mathrm{k}2\mathrm{\pi }, \mathrm{k}\in \mathrm{Z} \\ <=>\mathrm{x}=–\frac{\mathrm{\pi }}{12}+\mathrm{k\pi } \mathrm{hoặc} \mathrm{x}=\frac{7\mathrm{\pi }}{12}+\mathrm{k\pi }, \mathrm{k}\in \mathrm{Z} \\ \mathrm{t}=\frac{1}{3}<=>\sin 2\mathrm{x}=\frac{1}{3}<=>2\mathrm{x}=\arcsin \frac{1}{3}+\mathrm{k}2\mathrm{\pi } \mathrm{hoặc} 2\mathrm{x}=\mathrm{\pi }–\arcsin \frac{1}{3}+\mathrm{k}2\mathrm{\pi }, \mathrm{k}\in \mathrm{Z} \\ <=>\mathrm{x}=\frac{1}{2}\arcsin \left(\frac{1}{3}\right)+\mathrm{k\pi } \mathrm{hoặc} \mathrm{x}=\frac{\mathrm{\pi }}{2}–\frac{1}{2}\arcsin \left(\frac{1}{3}\right)+\mathrm{k\pi }, \mathrm{k}\in \mathrm{Z} \\ \mathrm{Vậy} \mathrm{S}=\left\{\frac{–\mathrm{\pi }}{12}+\mathrm{k\pi };\frac{7\mathrm{\pi }}{12}+\mathrm{k\pi };\frac{1}{2}\arcsin \left(\frac{1}{3}\right)+\mathrm{k\pi };\frac{\mathrm{\pi }}{2}–\frac{1}{2}\arcsin \left(\frac{1}{3}\right)+\mathrm{k\pi }\right\}, \mathrm{k}\in \mathrm{Z} \end{gathered}$$

   $$\begin{gathered} {\sin }^{4}x–{\cos }^{4}x=\frac{1}{2} \\ <=>\left({\sin }^{2}x–{\cos }^{2}x\right).\left({\sin }^{2}x+{\cos }^{2}x\right)=\frac{1}{2} \\ <=>–\cos 2x=\frac{1}{2} \\ <=>\cos 2x=–\frac{1}{2} \\ <=>2x=\pm \frac{2\mathrm{\pi }}{3}+k2\mathrm{\pi }, \mathrm{k}\in \mathrm{Z} \\ <=>\mathrm{x}=\pm \frac{\mathrm{\pi }}{3}+\mathrm{k\pi }, \mathrm{k}\in \mathrm{Z} \\ \mathrm{Vậy} \mathrm{x}=\pm \frac{\mathrm{\pi }}{3}+\mathrm{k\pi }, \mathrm{k}\in \mathrm{Z} \end{gathered}$$

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