1) $y=0$
$\Rightarrow \sin x-m\cos x-m-2=0$
$\Rightarrow \sin x-m\cos x=m+2$
$\Rightarrow \dfrac{1}{\sqrt{1+m^2}}\sin x-\dfrac{m}{\sqrt{1+m^2}}\cos x=\dfrac{m+2}{\sqrt{1+m^2}}$
Đặt $\dfrac{1}{\sqrt{1+m^2}}=\cos \alpha$ và $\dfrac{m}{\sqrt{1+m^2}}=\sin\alpha$
Phương trình $\Rightarrow\cos\alpha\sin x-\sin\alpha\cos x=\dfrac{m+2}{\sqrt{1+m^2}}$
$\Rightarrow \sin(x-\alpha)=\dfrac{m+2}{\sqrt{1+m^2}}$
Ta có: $-1\le\sin x\le1$ $\forall x$
$\Rightarrow -1\le\sin(x-\alpha)\le1$
$\Rightarrow -1\le\dfrac{m+2}{\sqrt{1+m^2}}\le1$
+) $\dfrac{m+2}{\sqrt{1+m^2}}\le1$
$\Rightarrow \sqrt{1+m^2}\ge m+2$
$\Rightarrow \left[\begin{array}{l} \left\{ \begin{array}{l} 1+m^2\ge0 \\ m+2\le0\end{array} \right . \\\left\{ \begin{array}{l} m+2>0 \\ 1+m^2\ge(m+2)^2\end{array} \right .\end{array} \right .$
$\Rightarrow \left[\begin{array}{l} m\le-2\\\left\{ \begin{array}{l} m>-2 \\ m\le\dfrac{-3}{4}\end{array} \right .\end{array} \right .$
$\Rightarrow m\le\dfrac{-3}{4}$
+) $\dfrac{m+2}{\sqrt{1+m^2}}\ge-1$
$\Rightarrow \sqrt{1+m^2\ge-m-2}$
$\Rightarrow \left[\begin{array}{l} \left\{ \begin{array}{l} 1+m^2\ge0 \\ -m-2\le0\end{array} \right . \\\left\{ \begin{array}{l} -m-2>0 \\ 1+m^2\ge(-m-2)^2\end{array} \right .\end{array} \right .$
$\Rightarrow \left[\begin{array}{l} m\ge-2\\\left\{ \begin{array}{l} m<-2 \\ m\le\dfrac{-3}{4}\end{array} \right .\end{array} \right .$
$\Rightarrow -2\le m\le\dfrac{-3}{4}$
$\Rightarrow \left\{ \begin{array}{l} m\ge-2 \\ m\le-2 \end{array} \right .$
$\Rightarrow \forall m$
Vậy kết hợp hai điều kiện $\Rightarrow m\le\dfrac{-3}{4}$.
