a) y= -sin2x+cosx+2x-3
⇒y'= -2.cos2x-sinx+2
ta có: y'=0
⇔ -2cos2x-sinx+2=0
⇔ -2(1-2sin²x)-sinx+2=0
⇔ -2+4sin²x-sinx+2=0
⇔ 4sin²x-sinx=0
⇔sinx(4sinx-1)=0
⇔\(\left[ \begin{array}{l}sinx=0\\4sinx-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=kπ\\sinx=\frac{1}{4} \end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=kπ\\x=arc.sin\frac{1}{4}+k2π\\x=π-arc.sin\frac{1}{4}+k2π \end{array} \right.\)
b)y=$\sqrt{4-x²}$
⇒y' = $\frac{(4-x²)'}{2.\sqrt{4-x²}}$
⇔y'=$\frac{-2x}{2.\sqrt{4-x²}}$
⇔y'=$\frac{-x}{\sqrt{4-x²}}$
ta có: y'=0
⇒ x=0
