cosx – sinx/2 +1 = 0 giải bài này giúp em với ạ…em cảm ơn ạ❤️ 29/09/2021 Bởi Margaret cosx – sinx/2 +1 = 0 giải bài này giúp em với ạ…em cảm ơn ạ❤️
$\cos x – \sin\dfrac{x}{2} +1 = 0 $ $\Rightarrow 1-2{\sin}^2\dfrac{x}{2}- \sin\dfrac{x}{2} +1 = 0$ $\Rightarrow 2{\sin}^2\dfrac{x}{2}+\sin\dfrac{x}{2} -2 = 0$ $\Rightarrow\left[ \begin{array}{l} \sin\dfrac{x}{2}=\dfrac{-1+\sqrt{17}}{4} \\ \sin\dfrac{x}{2}=\dfrac{-1-\sqrt{17}}{4}(<-1)\text{ (loại)}\end{array} \right .$ $\Rightarrow\left[ \begin{array}{l} \dfrac{x}{2}=\arcsin\dfrac{-1+\sqrt{17}}{4} +k2\pi\\ \dfrac{x}{2}=\pi-\arcsin\dfrac{-1+\sqrt{17}}{4}+k2\pi\end{array} \right .$ $\Rightarrow\left[ \begin{array}{l} x=2\arcsin\dfrac{-1+\sqrt{17}}{4} +k4\pi\\ x=2\pi-2\arcsin\dfrac{-1+\sqrt{17}}{4}+k4\pi\end{array} \right .(k\in\mathbb Z)$ Bình luận
$\cos x – \sin\dfrac{x}{2} +1 = 0 $
$\Rightarrow 1-2{\sin}^2\dfrac{x}{2}- \sin\dfrac{x}{2} +1 = 0$
$\Rightarrow 2{\sin}^2\dfrac{x}{2}+\sin\dfrac{x}{2} -2 = 0$
$\Rightarrow\left[ \begin{array}{l} \sin\dfrac{x}{2}=\dfrac{-1+\sqrt{17}}{4} \\ \sin\dfrac{x}{2}=\dfrac{-1-\sqrt{17}}{4}(<-1)\text{ (loại)}\end{array} \right .$ $\Rightarrow\left[ \begin{array}{l} \dfrac{x}{2}=\arcsin\dfrac{-1+\sqrt{17}}{4} +k2\pi\\ \dfrac{x}{2}=\pi-\arcsin\dfrac{-1+\sqrt{17}}{4}+k2\pi\end{array} \right .$ $\Rightarrow\left[ \begin{array}{l} x=2\arcsin\dfrac{-1+\sqrt{17}}{4} +k4\pi\\ x=2\pi-2\arcsin\dfrac{-1+\sqrt{17}}{4}+k4\pi\end{array} \right .(k\in\mathbb Z)$