Đáp án: $ab=-7$
Giải thích các bước giải:
Ta có:
$\cos2x+\sin^2x+3\cos x-m=5$
$\to(\cos^2x-\sin^2x)+\sin^2x+3\cos x-m=5$
$\to\cos^2x+3\cos x-m=5$
$\to\cos^2x+3\cos x-5=m$
$\to (\cos x+\dfrac32)^2-\dfrac{29}4=m$
Mà $-1\le \cos x\le 1$
$\to \dfrac12\le \cos x+\dfrac32\le \dfrac52$
$\to \dfrac14\le (\cos x+\dfrac32)^2\le \dfrac{25}4$
$\to -7\le (\cos x+\dfrac32)^2-\dfrac{29}4\le -1$
$\to -7\le m\le -1$
$\to m\in[-7,-1]$
$\to a=-7, b=-1$
$\to ab=-7$
