Bài `2:`
`1)y=sin^2 x+2sin x cos x-3cos^2 x+5`
`<=>y=[1-cos 2x]/2+sin 2x-3[1+cos 2x]/2+5`
`<=>y=1/2(1-cos 2x+2sin 2x-3-3cos 2x+10)`
`<=>y=1/2(2sin 2x-4cos 2x)+4`
`<=>y=\sqrt{5}(2/[2\sqrt{5}]sin 2x-4/[2\sqrt{5}]cos 2x)+4`
Đặt `2/[2\sqrt{5}]=cos \alpha;4/[2\sqrt{5}]=sin \alpha`
`=>y=\sqrt{5}sin(2x-\alpha)+4`
Có: `-1 <= sin (2x-\alpha) <= 1 AA x`
`<=>-\sqrt{5} <= \sqrt{5}sin(2x-\alpha) <= \sqrt{5}`
`<=>-\sqrt{5}+4 <= y <= \sqrt{5}+4`
`=>y_[mi n]=-\sqrt{5}+4<=>sin(2x-\alpha)=-1<=>2x-\alpha=-\pi/2+k2\pi`
`<=>x=[\alpha]/2-\pi/4+k\pi` `(k in ZZ)`
`y_[max]=\sqrt{5}+4<=>sin(2x-\alpha)=1<=>2x-\alpha=\pi/2+k2\pi`
`<=>x=[\alpha]/2+\pi/4+k\pi` `(k in ZZ)`
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`2)y=cos 2x+5sin x+2`
`<=>y=1-2sin^2 x+5sin x+2`
`<=>y=-2sin^2 x+5sin x+3`
Đặt `sin x=t` `(t in [-1;1])`
`=>y=-2t^2+5t+3` có `I(5/4;49/8)`
BBT:
`=>y_[mi n]=-4<=>sin x=-1<=>x=-\pi/2+k2\pi` `(k in ZZ)`
`y_[max]=6<=>sin x=1<=>S=\pi/2+k2\pi` `(k in ZZ)`
