Đáp án:
$P_{\min}=-\dfrac{1}{3}$
Giải thích các bước giải:
$P=3\cos^2x+4\cos x+1$
Đặt $t=\cos x,\,t\in[-1;1]$
$P=3t^2+4t+1$
$=3\left(t^2+\dfrac{4}{3}t+\dfrac{1}{3}\right)$
$=3\left(t^2+2.t.\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{1}{9}\right)$
$=3\left(t+\dfrac{2}{3}\right)^2-\dfrac{1}{3}$
Ta có:
$\left(t+\dfrac{2}{3}\right)^2\ge 0$
$⇒3\left(t+\dfrac{2}{3}\right)^2\ge 0$
$⇒3\left(t+\dfrac{2}{3}\right)^2-\dfrac{1}{3}\ge -\dfrac{1}{3}$
$⇒P\ge -\dfrac{1}{3}⇒P_{\min}=-\dfrac{1}{3}$
Dấu "=" xảy ra khi:
$\left(t+\dfrac{2}{3}\right)^2=0$
$⇒t+\dfrac{2}{3}=0$
$⇒t=-\dfrac{2}{3}\,(TM)$
$⇒\cos x=-\dfrac{2}{3}$
$⇒x=\arccos\left(-\dfrac{2}{3}\right)$
Vậy $P_{\min}=-\dfrac{1}{3}$ khi $x=\arccos\left(-\dfrac{2}{3}\right)$.
