a.

\(y=sinx.cosx+1=\dfrac{1}{2}sin2x+1\)

\(-1\le sin2x\le1\Rightarrow\dfrac{1}{2}\le y\le\dfrac{3}{2}\)

\(y_{min}=\dfrac{1}{2}\) khi \(sin2x=-1\Rightarrow x=-\dfrac{\pi}{4}+k\pi\)

\(y_{max}=\dfrac{3}{2}\) khi \(sin2x=1\Rightarrow x=\dfrac{\pi}{4}+k\pi\)

b.

\(y=2\left(\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx\right)-2=2.sin\left(x-\dfrac{\pi}{6}\right)-2\)

\(-1\le sin\left(x-\dfrac{\pi}{6}\right)\le1\Rightarrow-4\le y\le0\)

\(y_{min}=-4\) khi \(sin\left(x-\dfrac{\pi}{6}\right)=-1\Rightarrow x=-\dfrac{\pi}{3}+k2\pi\)

\(y_{max}=0\) khi \(sin\left(x-\dfrac{\pi}{6}\right)=1\Rightarrow x=\dfrac{2\pi}{3}+k2\pi\)

a) \(y=1-2sinx\)

Ta có: \(-1\le sinx\le1\Rightarrow-2\le2sinx\le2\)

                                   \(\Rightarrow2\ge-2sin2x\ge-2\)

                                   \(\Rightarrow3\ge1-2sinx\ge-1\)

      Vậy \(y_{max}=3,y_{min}=-1\)

a) làm tương tự 2 bài mk đã giải nha.

b) \(y=2\cos^2x-2\sqrt{3}\sin x\cos x+1\)

\(=1-\left(\cos2x+\sqrt{3}\sin2x\right)\)

Lại có \(-2\le\cos2x+\sqrt{3}\sin2x\le2\) \(\Rightarrow-1\le y\le3\)

c) Vì \(\left\{{}\begin{matrix}0\le\sqrt[4]{\sin x}\le1\\0\le\sqrt{\cos x}\le1\end{matrix}\right.\)

Do đó \(-1\le y\le1\)

a) Đặt \(t=sinx+cosx\)

\(\Rightarrow t^2=\left(sinx+cosx\right)^2\overset{bunhiacopxki}{\le}\left(1^2+1^2\right)\left(sinx^2+cosx^2\right)=2\\ \Rightarrow-\sqrt{2}\le t\le\sqrt{2}\\ \Rightarrow-\sqrt{2}+1\le y=t+1\le\sqrt{2}+1\)

Vậy \(Min\text{ }y=-\sqrt{2}+1\Leftrightarrow sinx=cosx=\frac{-1}{\sqrt{2}}\Leftrightarrow x=\pm\frac{3\pi}{4}+k2\pi\)

\(Max\text{ }y=\sqrt{2}+1\Leftrightarrow sinx=cosx=\frac{1}{\sqrt{2}}\Leftrightarrow x=\pm\frac{\pi}{4}+k2\pi\)

\(b\text{) }y=cosx-cos2x+4\\ =cosx-\left(2cos^2x-1\right)+4\\ =-2cos^2x+cosx+5\)

\(\Rightarrow Min\text{ }y=2\Leftrightarrow cosx=-1\)

\(\Rightarrow Max\text{ }y=\frac{41}{8}\Leftrightarrow cosx=\frac{1}{4}\)

\(\text{c) }y=2sin^2x+4\sqrt{3}sinx\cdot cosx+6cos^2x+1\\ =\left(1-cos2x\right)+2\sqrt{3}sin2x+3\left(cos2x+1\right)+1\\ =2cos2x+2\sqrt{3}sin2x+5\)

Đặt \(t=2cos2x+2\sqrt{3}sin2x\)

\(\Rightarrow t^2\le\left[2^2+\left(2\sqrt{3}\right)^2\right]\left(cos^22x+sin^22x\right)=16\\ \Rightarrow-4\le t\le4\\ \Rightarrow1\le y\le9\\ \)

Vậy \(Min\text{ }y=1\Leftrightarrow sin2x=-\frac{1}{2}\)

\(Max\text{ }y=9\Leftrightarrow sin2x=\frac{1}{2}\)

Chọn B

Chọn A

a.

\(-1\le sinx\le1\Rightarrow-7\le y\le-3\)

\(y_{min}=-7\) khi \(sinx=-1\)

\(y_{max}=-3\) khi \(sinx=1\)

b.

\(-1\le cos\left(x+\frac{\pi}{3}\right)\le1\Rightarrow1\le y\le5\)

\(y_{min}=1\) khi \(cos\left(x+\frac{\pi}{3}\right)=-1\)

\(y_{max}=5\) khi \(cos\left(x+\frac{\pi}{3}\right)=1\)

c.

\(0\le1-cosx\le2\Rightarrow-5\le y\le3\sqrt{2}-5\)

\(y_{min}=-5\) khi \(cosx=1\)

\(y_{max}=3\sqrt{2}-5\) khi \(cosx=-1\)

d.

ĐKXĐ: \(0\le sinx\Rightarrow0\le sinx\le1\Rightarrow1\le y\le3\)

\(y_{min}=1\) khi \(sinx=0\)

\(y_{max}=3\) khi \(sinx=1\)