a: -1<=sin x<=1

=>-1+3<=sin x+3<=1+3

=>2<=sinx+3<=4

=>\(\dfrac{1}{2}>=\dfrac{1}{sinx+3}>=\dfrac{1}{4}\)

=>\(2>=\dfrac{4}{sinx+3}>=1\)

=>\(-2< =-\dfrac{4}{sinx+3}< =-1\)

=>-2+3<=y<=-1+3

=>1<=y<=2

y=1 khi \(\dfrac{-4}{sinx+3}+3=1\)

=>\(\dfrac{-4}{sinx+3}=-2\)

=>sinx+3=2

=>sin x=-1

=>x=-pi/2+k2pi

y=3 khi sin x=1

=>x=pi/2+k2pi

b: -1<=cosx<=1

=>4>=-4cosx>=-4

=>9>=-4cosx+5>=1

=>2/9<=2/5-4cosx<=2

=>2/9<=y<=2

\(y_{min}=\dfrac{2}{9}\) khi \(\dfrac{2}{5-4cosx}=\dfrac{2}{9}\)

=>\(5-4\cdot cosx=9\)

=>4*cosx=4

=>cosx=1

=>x=k2pi

y max khi cosx=-1

=>x=pi+k2pi

c: \(0< =cos^2x< =1\)

=>\(0< =2\cdot cos^2x< =2\)

=>\(-1< =y< =2\)

y min=-1 khi cos^2x=0

=>x=pi/2+kpi

y max=2 khi cos^2x=1

=>sin^2x=0

=>x=kpi

\(y=2cos^2x-2\sqrt{3}sinx.cosx+1\)

\(=2cos^2x-1-2\sqrt{3}sinx.cosx+2\)

\(=cos2x-\sqrt{3}sin2x+2\)

\(=2\left(\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x\right)+2\)

\(=2cos\left(2x+\dfrac{\pi}{3}\right)+2\)

Ta có: \(cos\left(2x+\dfrac{\pi}{3}\right)\in\left[-1;1\right]\)

\(\Rightarrow min=0\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=-1\Leftrightarrow2x+\dfrac{\pi}{3}=\pi+k2\pi\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)

\(\Rightarrow max=4\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=1\Leftrightarrow2x+\dfrac{\pi}{3}=k2\pi\Leftrightarrow x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

\(y=2cos^2x-\sqrt{3}sin2x+1=cos2x-\sqrt{3}sin2x+2\)

\(y=2.cos\left(2x+\dfrac{\pi}{3}\right)+2\)

\(\forall x\in R->-1\le cos\left(2x+\dfrac{\pi}{3}\right)\)

=> \(Min_y=2.\left(-1\right)+2=0\) 

Mặt khác, theo Bunhiacopxki:

\(\left(cos2x+\sqrt{3}sin2x\right)^2\le\left(1^2+\sqrt{3}^2\right)\left(cos^22x+sin^22x\right)=4\)

=>\(Max_y=4\)

Đật 3 cái mẫu bên VT lần lượt là x,y,z rồi áp dụng C-S dạng engel

Để dễ nhìn ta đặt \(\hept{\begin{cases}\sqrt{2x-3}=a\\\sqrt{y-2}=b\\\sqrt{3z-1}=c\end{cases}\left(a,b,c\ge0\right)}\)

Vậy BĐT đầu tương đương \(T=\frac{1}{a}+\frac{4}{b}+\frac{16}{c}+a+b+c\)

Áp dụng BĐT C-S dạng Engel ta có:

\(\frac{1}{a}+\frac{4}{b}+\frac{16}{c}=\frac{1^2}{a}+\frac{2^2}{b}+\frac{4^2}{c}\ge\frac{\left(1+2+4\right)^2}{a+b+c}=\frac{49}{a+b+c}\)

Tiếp tục dùng AM-GM ta có: \(VT\ge\frac{49}{a+b+c}+\left(a+b+c\right)\ge2\sqrt{\frac{49}{a+b+c}\cdot\left(a+b+c\right)}=2\sqrt{49}=14\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}a=1\\b=2\\c=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=\frac{17}{3}\end{cases}}\)

nhìn qua thì chắc AM-GM+Cauchy-schwarz chắc thế :)

\(P=4sin^2x+\sqrt{2}\left(sin2x.cos\frac{\pi}{4}+cos2x.sin\frac{\pi}{4}\right)\)

\(P=4sin^2x+sin2x+cos2x\)

\(P=2\left(1-cos2x\right)+sin2x+cos2x\)

\(P=2+sin2x-cos2x\)

\(P=2+\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)\)

Do \(sin\left(2x-\frac{\pi}{4}\right)\le1\Rightarrow P\le2+\sqrt{2}\)

\(\Rightarrow P_{max}=2+\sqrt{2}\) khi \(sin\left(2x-\frac{\pi}{4}\right)=1\Leftrightarrow x=\frac{3\pi}{8}+k\pi\)