a,
$\sin x-m+1=0$
$\to \sin x=m-1$
Để PT có nghiệm:
$-1\le m-1\le 1$
$\to 0\le m\le 2$
b,
$2\cos\dfrac{x}{2}+m-4=0$
$\to \cos\dfrac{x}{2}+\dfrac{-m+4}{2}$
Để PT có nghiệm:
$-1\le \dfrac{-m+4}{2}\le 1$
$\to -2\le -m+4\le 2$
$\to -6\le -m\le -2$
$\to 2\le m\le 6$
c,
$m\cos x-1=(2m+1)\cos x$
$\to (m+1)\cos x=-1$
ĐK: $m+1\ne 0\to m\ne -1$
Để PT có nghiệm:
$-1\le \dfrac{-1}{m+1}\le 1$
$\dfrac{-1}{m+1}\ge -1\to \dfrac{-1}{m+1}+1\ge 0\to \dfrac{m}{m+1}\ge 0\to m\in (-\infty;-1)\cup[0;+\infty)$
$\dfrac{-1}{m+1}\le 1\to \dfrac{-1}{m+1}-1\le 0\to \dfrac{-m-2}{m+1}\le 0\to \dfrac{m+2}{m+1}\ge 0\to m\in (-\infty;-2]\cup(-1;+\infty)$
Vậy $m\in (-\infty;-2]\cup[0;+\infty)$
d,
$\sin^2x-2\cos x=m$
$\to 1-\cos^2x-2\cos x=m$
$\to cos^2x+2\cos x+m-1=0$
Đặt $t=\cos x$
$\to t^2+2t+m-1=0$ (*)
Để PT có nghiệm thì nghiệm của (*) thuộc $[-1;1]$
$\Delta'= 1-m+1= m$
ĐK: $m\ge 0$
$\to t_1=-1+\sqrt{m}; t_2=-1-\sqrt{m}$
$-1\le -1+\sqrt{m}\le 1\to 0\le \sqrt{m}\le 2\to 0\le m\le 4$
$-1\le -1-\sqrt{m}\le 1\to 0\le -\sqrt{m}\le 2\to -2\le \sqrt{m}\le 0\to m=0$
Vậy $m=0$
