Question

Answer 1

\(y=\dfrac{2\cos x+3}{\cos x-2}=\dfrac{2\left(\cos x-2\right)+7}{\cos x-2}=2+\dfrac{7}{\cos x-2}\)

Với \(-1\le\cos x\le1\) thì:

\(2+\dfrac{7}{1-2}\le y\le2+\dfrac{7}{-1-2}\Leftrightarrow-5\le y\le-\dfrac{1}{3}\)

Dấu "=" xảy ra tại min max: 

\(x=\pi+k2\pi\left(min\right)\left(k\in Z\right)\) và \(x=l2\pi\left(max\right)\left(l\in Z\right)\)