Giải phương trình: 2Cos 2$Cos^{2}$ x = 3$Sin^{2}$ 5x + 2 25/09/2021 Bởi Liliana Giải phương trình: 2Cos 2$Cos^{2}$ x = 3$Sin^{2}$ 5x + 2
$2{\cos}^2x=3{\sin}^25x+2$ $\Rightarrow 2(1-{\sin}^2x)=3{\sin}^25x+2$ $\Rightarrow2{\sin}^2x+3{\sin}^25x=0 $ Mà $2{\sin}^2x+3{\sin}^25x\ge0$ $\forall x$ Khi đó $2{\sin}^2x+3{\sin}^25x=0 $ $\Leftrightarrow \sin x=\sin 5x=0$ $\Rightarrow \left\{ \begin{array}{l}x=k\pi\\ 5x=k\pi \end{array} \right .$ $\Rightarrow \left\{ \begin{array}{l}x=k\pi\\x=k\dfrac{\pi}{5} \end{array} \right .$ $\Rightarrow k\pi=k\dfrac{\pi}{5}$ $\Rightarrow k=0$ $\Rightarrow x=0$ Bình luận
$2{\cos}^2x=3{\sin}^25x+2$
$\Rightarrow 2(1-{\sin}^2x)=3{\sin}^25x+2$
$\Rightarrow2{\sin}^2x+3{\sin}^25x=0 $
Mà $2{\sin}^2x+3{\sin}^25x\ge0$ $\forall x$
Khi đó $2{\sin}^2x+3{\sin}^25x=0 $
$\Leftrightarrow \sin x=\sin 5x=0$
$\Rightarrow \left\{ \begin{array}{l}x=k\pi\\ 5x=k\pi \end{array} \right .$
$\Rightarrow \left\{ \begin{array}{l}x=k\pi\\x=k\dfrac{\pi}{5} \end{array} \right .$
$\Rightarrow k\pi=k\dfrac{\pi}{5}$
$\Rightarrow k=0$
$\Rightarrow x=0$