Vì $-1 \le x \le 1$. Đặt $x = \cos t, \ t \in[0,\ \pi]$ thay vào ta có$\sqrt{1 - \cos^2 t} = 4\cos^3 t - 3\cos t$$\Leftrightarrow \sin t = \cos 3t$$\Leftrightarrow \cos (\dfrac{\pi}{2} - t ) = \cos 3t$$\Leftrightarrow \left [ \begin{matrix} t = -\dfrac{3\pi}{8} + k\pi \\ t = -\dfrac{3\pi}{8} + k\pi \\ t = -\dfrac{3\pi}{4} + k\pi \end{matrix} \right. \ \ \ \ \quad \quad k \in Z$Vì $t \in [0,\ \pi] \Rightarrow t = \dfrac{3\pi}{4},\ \dfrac{\pi}{8},\ \dfrac{5\pi}{8}$Vậy $x = \cos \dfrac{3\pi}{4} = -\dfrac{1}{\sqrt 2}, x = \cos \dfrac{\pi}{8}, \ x = \cos \dfrac{5\pi}{8}$
Vì $-1 \le x \le 1$. Đặt $x = \cos t$ thay vào ta có$\sqrt{1 - \cos^2 t} = 4\cos^3 t - 3\cos t$$\Leftrightarrow \sin t = \cos 3t$$\Leftrightarrow \cos (\dfrac{\pi}{2} - t ) = \cos 3t$Bạn tự làm nốt
Vì $-1 \le x \le 1$. Đặt $x = \cos t
, \ t \in[0,\ \pi]$ thay vào ta có$\sqrt{1 - \cos^2 t} = 4\cos^3 t - 3\cos t$$\Leftrightarrow \sin t = \cos 3t$$\Leftrightarrow \cos (\dfrac{\pi}{2} - t ) = \cos 3t$
$\Leftrightarrow \left [ \begin
{matrix} t
= -\dfrac{3\pi}{8} + k\pi \\ t = -\dfrac{3\pi}{8} + k\pi \\ t = -\dfrac{3\pi}{4} + k\pi \end{m
atrix} \right. \ \ \ \ \quad \quad k \in
Z$Vì $t
\in [0,\ \pi] \Rightarrow t = \dfrac{3\pi}{4},\ \dfrac{\pi}{8},\ \dfrac{5\pi}{8}$Vậy $x = \cos \dfrac{3\pi}{4} = -\dfrac{1}{\sqrt 2}, x = \cos \dfrac{\pi}{8}, \ x = \cos \dfrac{5\pi}{8}$