1.

\(sin\left(x-\dfrac{\pi}{3}\right)\in\left[-1;1\right]\)

\(\Rightarrow y=2sin\left(x-\dfrac{\pi}{3}\right)+3\in\left[1;5\right]\)

\(\Rightarrow\left\{{}\begin{matrix}y_{min}=1\\y_{max}=5\end{matrix}\right.\)

2.

\(cos2x\in\left[-1;1\right]\)

\(\Rightarrow y=3-\dfrac{1}{2}cos2x\in\left[\dfrac{5}{2};\dfrac{7}{2}\right]\)

\(\Rightarrow\left\{{}\begin{matrix}y_{min}=\dfrac{5}{2}\\y_{max}=\dfrac{7}{2}\end{matrix}\right.\)

Hàm số xác định khi:

\(\left\{{}\begin{matrix}\dfrac{1-sinx}{2+cosx}\ge0\\2+cosx\ne0\\1-cosx>0\end{matrix}\right.\Leftrightarrow1-cosx\ne0\Leftrightarrow x\ne k2\pi\)

TXĐ: D=R

\(y=\dfrac{1}{2}+\dfrac{1}{2}cos2x-1=\dfrac{1}{2}cos2x-\dfrac{1}{2}\)

Với mọi \(x\in R\Rightarrow x+\pi\in R\)

\(f\left(x+\pi\right)=\dfrac{1}{2}cos\left(2x+2\pi\right)-\dfrac{1}{2}=\dfrac{1}{2}cos2x-\dfrac{1}{2}=f\left(x\right)\)

\(\Rightarrow\) Hàm là hàm tuần hoàn với chu kì \(T=\pi\)

\(y=f\left(x\right)=\dfrac{cosx}{2}\)

\(f\left(-x\right)=\dfrac{cos\left(-x\right)}{2}=\dfrac{cosx}{2}=f\left(x\right)\)

\(\Rightarrow\) Là hàm số chẵn.

Khi khai triển ngoặc thứ 2, ta có \(2.1=4?\)

U là tr, 

\(y=f\left(x\right)=\sqrt{5-2sin^2x.cos^2x}=\sqrt{5-\dfrac{1}{2}sin^22x}\)

\(f\left(-x\right)=\sqrt{5-\dfrac{1}{2}sin^2\left(-2x\right)}=\sqrt{5-\dfrac{1}{2}sin^22x}=f\left(x\right)\)

\(\Rightarrow\) Là hàm số chẵn.

Hàm số nào vậy em?

c, \(sin\alpha=-1\Leftrightarrow\alpha=-\dfrac{\pi}{2}+k2\pi\)

a, \(sin\alpha=0\Leftrightarrow\alpha=k\pi\)

\(A=cosx+cos3x+cos5x+cos7x\)

\(=2cos4x.cos3x+2cos4x.cosx\)

\(=2cos4x.\left(cos3x+cosx\right)\)

\(=4cos4x.cos2x.cosx\)