sin^6x + cos^6x = 4cos^2(2x)

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

$\begin{cases}2x=\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k2\pi\\2x=\pi-\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k2\pi\end{cases}$

$\Leftrightarrow\begin{cases}x=\dfrac12\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k\pi\\x=\dfrac{\pi}2-\dfrac12\arcsin± \dfrac{2\sqrt3}{\sqrt{13}}+k2\pi\end{cases}$ $(k\in\mathbb Z)$