Đáp án:
$\tan a=-\dfrac{5}{12}$
Giải thích các bước giải:
ĐKXĐ: $\cos a\ne 0$
$2\sin a+3\cos a=2$
$⇒\sin a=\dfrac{2-3\cos a}{2}$
Ta có:
$\sin^2a+\cos^2a=1$
$⇒\left(\dfrac{2-3\cos a}{2}\right)^2+\cos^2a=1$
$⇒\dfrac{(2-3\cos a)^2}{4}+\dfrac{4\cos^2a}{4}=\dfrac{4}{4}$
$⇒4-12\cos a+9\cos^2a+4\cos^2a=4$
$⇒13\cos^2a-12\cos a=0$
$⇒\cos a(13\cos a-12)=0$
$⇒\left[ \begin{array}{l}\cos a=0\\13\cos a-12=0\end{array} \right.⇒\left[ \begin{array}{l}\cos a=0\,(L)\\\cos a=\dfrac{12}{13}\end{array} \right.$
$⇒\sin a=\dfrac{2-3.\dfrac{12}{13}}{2}=-\dfrac{5}{13}$
$⇒\tan a=\dfrac{\sin a}{\cos a}=\dfrac{-\dfrac{5}{13}}{\dfrac{12}{13}}=-\dfrac{5}{12}$
Vậy $\tan a=-\dfrac{5}{12}$.
