Đáp án: $\begin{cases}x=\dfrac12\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k\pi\\x=\dfrac{\pi}2-\dfrac12\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k2\pi\end{cases}$ $(k\in\mathbb Z)$ Lời giải: $\sin^6x+\cos^6x=4\cos²x$ $\Leftrightarrow(\sin²x+\cos²x)^3-3\sin²x\cos²x=4\cos²x$ $\Leftrightarrow1-\dfrac34\sin²2x=4-4\sin²2x$ $\Leftrightarrow\dfrac{13}4\sin²2x-3=0$ $\Leftrightarrow\sin²2x=\dfrac{12}{13}$ $\Leftrightarrow\sin2x=± \dfrac{2\sqrt3}{\sqrt{13}}$ Vậy phương trình có nghiệm $\begin{cases}2x=\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k2\pi\\2x=\pi-\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k2\pi\end{cases}$ $\Leftrightarrow\begin{cases}x=\dfrac12\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k\pi\\x=\dfrac{\pi}2-\dfrac12\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k2\pi\end{cases}$ $(k\in\mathbb Z)$ Bình luận
Đáp án:
$\begin{cases}x=\dfrac12\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k\pi\\x=\dfrac{\pi}2-\dfrac12\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k2\pi\end{cases}$ $(k\in\mathbb Z)$
Lời giải:
$\sin^6x+\cos^6x=4\cos²x$
$\Leftrightarrow(\sin²x+\cos²x)^3-3\sin²x\cos²x=4\cos²x$
$\Leftrightarrow1-\dfrac34\sin²2x=4-4\sin²2x$
$\Leftrightarrow\dfrac{13}4\sin²2x-3=0$
$\Leftrightarrow\sin²2x=\dfrac{12}{13}$
$\Leftrightarrow\sin2x=± \dfrac{2\sqrt3}{\sqrt{13}}$
Vậy phương trình có nghiệm
$\begin{cases}2x=\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k2\pi\\2x=\pi-\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k2\pi\end{cases}$
$\Leftrightarrow\begin{cases}x=\dfrac12\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k\pi\\x=\dfrac{\pi}2-\dfrac12\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k2\pi\end{cases}$ $(k\in\mathbb Z)$