Đáp án:
$Q\in\left\{-\dfrac{7}{5};-\dfrac{1}{5};\dfrac{1}{5};\dfrac{7}{5}\right\}$
Giải thích các bước giải:
$\tan2x=\dfrac{24}{7}$
$⇒\dfrac{2\tan x}{1-\tan^2x}=\dfrac{24}{7}$
$⇒14\tan x=24-24\tan^2x$
$⇒12\tan^2x+7\tan x-12=0$
$⇒\left[ \begin{array}{l}\tan x=\dfrac{3}{4}\\\tan x=-\dfrac{4}{3}\end{array} \right.⇒\left[ \begin{array}{l}\dfrac{\sin x}{\cos x}=\dfrac{3}{4}\\\dfrac{\sin x}{\cos x}=-\dfrac{4}{3}\end{array} \right.⇒\left[ \begin{array}{l}\cos x=\dfrac{4}{3}\sin x\\\cos x=-\dfrac{3}{4}\sin x\end{array} \right.$
TH1: $\cos x=\dfrac{4}{3}\sin x$
$⇒\sin^2x+\cos^2x=1$
$⇒\sin^2x+\dfrac{16}{9}\sin^2x=1$
$⇒\dfrac{25}{9}\sin^2x=1$
$⇒\sin^2x=\dfrac{9}{25}$
$⇒\sin x=±\dfrac{3}{5}$
$⇒\cos x=±\dfrac{4}{5}$
$⇒Q=\sin x-\cos x=\mp\dfrac{1}{5}$
TH2: $\cos x=-\dfrac{3}{4}\sin x$
$⇒\sin^2x+\cos^2x=1$
$⇒\sin^2x+\dfrac{9}{16}\sin^2x=1$
$⇒\dfrac{25}{16}\sin^2x=1$
$⇒\sin^2x=\dfrac{16}{25}$
$⇒\sin x=±\dfrac{4}{5}$
$⇒\cos x=\mp\dfrac{3}{5}$
$⇒Q=\sin x-\cos x=±\dfrac{7}{5}$
Vậy $Q\in\left\{-\dfrac{7}{5};-\dfrac{1}{5};\dfrac{1}{5};\dfrac{7}{5}\right\}$.
