Đáp án:
$m=1.$
Giải thích các bước giải:
$\sin^2x+\sin x.\cos x-2\cos^2x=m\\ \Leftrightarrow \sin^2x+\cos^2x+\sin x.\cos x-3\cos^2x=m\\ \Leftrightarrow 1+\dfrac{1}{2}.2\sin x.\cos x-3\cos^2x=m\\ \Leftrightarrow \dfrac{1}{2}\sin 2x-3.\dfrac{1+\cos 2x}{2}=m-1\\ \Leftrightarrow \sin 2x-3(1+\cos 2x)=2(m-1)\\ \Leftrightarrow \sin 2x-3-3\cos 2x=2m-2\\ \Leftrightarrow \sin 2x-3\cos 2x=2m+1(1)\\ x=\dfrac{\pi}{2} + k \pi(k \in \mathbb{Z})\\ (1) \Leftrightarrow \sin \left(\pi+k 2 \pi\right)-3\cos \left(\pi+k 2 \pi\right)=2m+1\\ \Leftrightarrow \sin \left(\pi\right)-3\cos \left(\pi\right)=2m+1\\ \Leftrightarrow 0-3.(-1)=2m+1\\ \Leftrightarrow 3=2m+1\\ \Leftrightarrow 2=2m\\ \Leftrightarrow m=1.$
