Question

tìm GTLN, GTNN

\(y=\sqrt{3}\sin2x+\cos2x\)

Answer 1

`@TXĐ: D=RR`

`@` Có: `y=\sqrt{3}sin 2x+cos 2x`

       `<=>y=2(\sqrt{3}/2sin 2x+1/2cos 2x)`

       `<=>y=2sin(2x+\pi/6)`

Vì `-1 <= sin (2x+\pi/6) <= 1`

`<=>-2 <= 2sin(2x+\pi/6) <= 2`

`<=>-2 <= y <= 2`

Vậy `Max _y=2<=>sin(2x+\pi/6)=1<=>2x+\pi/6=\pi/2+k2\pi` 

                                       `<=>x=\pi/6+k\pi`  `(k in ZZ)`

       `Mi n _y=-2<=>sin(2x+\pi/6)=-1<=>2x+\pi/6=-\pi/2+k2\pi` 

                                       `<=>x=-\pi/3+k\pi`  `(k in ZZ)`

Answer 2

\(y=\sqrt{3}sin2x+cos2x=\dfrac{1}{2}sin\left(2x+\dfrac{pi}{3}\right)\)

=>-1/2<=y<=1/2

y=-1/2 khi 2x+pi/3=-pi/2+k2pi

=>2x=-5/6pi+k2pi

=>x=-5/12pi+kpi

y=1/2 khi 2x+pi/3=pi/2+k2pi

=>x=pi/12+kpi