a) Tập xác định của hàm số là \(D = \mathbb{R}\)

Vì \( - 1 \le \sin \left( {x - \frac{\pi }{4}} \right) \le 1 \Rightarrow  - 2 \le 2\sin \left( {x - \frac{\pi }{4}} \right) \le 2\; \Rightarrow  - 2 - 1 \le 2\sin \left( {x - \frac{\pi }{4}} \right) - 1 \le 2 - 1\)

\( \Rightarrow  - 3 \le 2\sin \left( {x - \frac{\pi }{4}} \right) - 1 \le 1\)

Vây tập giá trị của hàm số \(y = 2\sin \left( {x - \frac{\pi }{4}} \right) - 1\) là \(T = \left[ { - 3;1} \right]\).

b) Tập xác định của hàm số là \(D = \mathbb{R}\)

Vì \( - 1 \le \cos x \le 1 \Rightarrow 0 \le 1 + \cos x \le 2 \Rightarrow 0 \le \sqrt {1 + \cos x}  \le \sqrt 2 \;\; \Rightarrow  - 2 \le \sqrt {1 + \cos x}  - 2 \le \sqrt 2  - 2\)

Vậy tập giá trị của hàm số \(y = \sqrt {1 + \cos x}  - 2\) là \(T = \left[ { - 2;\sqrt 2  - 2} \right]\)

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

 \(y=2-\left(-cosx\right).\left(-sinx\right)\)

y = 2 - sinx.cosx

y = \(2-\dfrac{1}{2}sin2x\)

Max = 2 + \(\dfrac{1}{2}\) = 2,5

Min = \(2-\dfrac{1}{2}\) = 1,5

2, y = \(\sqrt{5-\dfrac{1}{2}sin^22x}\)

Min = \(\sqrt{5-\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{2}\)

Max = \(\sqrt{5}\)

a: \(5-2\cdot cos^2x\cdot\sin^2x\)

\(=5-2\cdot\left(\sin x\cdot cosx\right)^2\)

\(=5-2\cdot\left(\frac12\cdot\sin2x\right)^2=5-2\cdot\frac14\cdot\sin^22x=-\frac12\cdot\sin^22x+5\)

Ta có: \(0\le\sin^22x\le1\)

=>\(-\frac12\le-\frac12\cdot\sin^22x\le0\)

=>\(-\frac12+5\le-\frac12\cdot\sin^22x+5\le0+5\)

=>\(\frac92\le-\frac12\cdot\sin^22x+5\le5\)

=>\(\frac{3\sqrt2}{2}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)

=>\(4:\frac{3\sqrt2}{2}\ge\frac{4}{\sqrt{-\frac12\cdot sin^22x+5}}\ge\frac{4}{\sqrt5}\)

=>\(\frac{2\sqrt2}{3}\ge y\ge\frac{4\sqrt5}{5}\)

Do đó: \(y_{\max}=\frac{2\sqrt2}{3}\) khi \(\sin^22x=1\)

=>\(cos^22x=0\)

=>cos2x=0

=>\(2x=\frac{\pi}{2}+k\pi\)

=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(y_{\min}=\frac{4\sqrt5}{5}\) khi \(\sin^22x=0\)

=>sin 2x=0

=>\(2x=k\pi\)

=>\(x=\frac{k\pi}{2}\)

b: \(f\left(x\right)=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos2x-2\)

\(=3\cdot\sin^2x+5\cdot cos^2x-4\left(cos^2x-\sin^2x\right)-2\)

\(=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos^2x+4\cdot\sin^2x-2\)

\(=7\cdot\sin^2x+cos^2x-2=7\cdot\sin^2x+1-\sin^2x-2=6\cdot\sin^2x-1\)

Ta có: \(0\le\sin^2x\le1\)

=>\(0\le6\sin^2x\le6\)

=>\(0-1\le6\sin^2x-1\le6-1\)

=>-1<=f(x)<=5

f(x) min=-1 khi \(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

f(x) max=5 khi \(\sin^2x=1\)

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

=>cosx=0

=>\(x=\frac{\pi}{2}+k\pi\)

Ta có:

a) \(\sin \left( {\alpha  + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} = \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} + \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{1}{2} = \frac{{ - \sqrt 3  + 3\sqrt 2 }}{6}\)      

b) \(\cos \left( {\alpha  + \frac{\pi }{6}} \right) = \cos \alpha .\cos \frac{\pi }{6} - \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 6 }}{3}.\frac{1}{2} =  - \frac{{3 + \sqrt 6 }}{6}\)

c) \(\sin \left( {\alpha  - \frac{\pi }{3}} \right) = \sin \alpha \cos \frac{\pi }{3} - \cos \alpha \sin \frac{\pi }{3} = \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} = \frac{{3 + \sqrt 6 }}{6}\)

d) \(\cos \left( {\alpha  - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)

1: Ta có: \(-1<=\sin\left(2x+\frac{\pi}{4}\right)\le1\)

=>\(-3\le3\cdot\sin\left(2x+\frac{\pi}{4}\right)\le3\)

=>\(-3-1\le3\cdot\sin\left(2x+\frac{\pi}{4}\right)-1\le3-1\)

=>-4<=y<=2

=>Tập giá trị là T=[-4;2]

\(y_{\min}=-4\) khi \(\sin\left(2x+\frac{\pi}{4}\right)=-1\)

=>\(2x+\frac{\pi}{4}=-\frac{\pi}{2}+k2\pi\)

=>\(2x=-\frac34\pi+k2\pi\)

=>\(x=-\frac38\pi+k\pi\)

2: \(0\le cos^2x\le1\)

=>\(0\ge-5\cdot cos^2x\ge-5\)

=>\(0+3\ge-5\cdot cos^2x+3\ge-5+3\)

=>3>=y>=-2

=>Tập giá trị là T=[-2;3]

\(y_{\max}=3\) khi \(cos^2x=1\)

=>\(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

\(y_{\min}=-2\) khi \(cos^2x=0\)

=>cosx=0

=>\(x=\frac{k\pi}{2}\)

3: \(-1\le cosx\le1\)

=>\(-3\le3\cdot cosx\le3\)

=>\(-3+4\le3\cdot cosx+4\le3+4\)

=>\(1\le3\cdot cosx+4\le7\)

=>\(\frac51\ge\frac{5}{3\cdot cosx+4}\ge\frac57\)

=>\(\frac57\le y\le5\)

=>Tập giá trị là \(T=\left\lbrack\frac57;5\right\rbrack\)

\(y_{\min}=\frac57\) khi cosx=1

=>\(x=k2\pi\)

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

=>\(x=\pi+k2\pi\)

4: \(y=\sin^2x-4\cdot\sin x+8\)

\(=\sin^2x-4\cdot\sin x+4+4\)

\(=\left(\sin x-2\right)^2+4\)

Ta có: \(-1\le\sin x\le1\)

=>\(-1-2\le\sin x-2\le1-2\)

=>\(-3\le\sin x-2\le-1\)

=>\(1\le\left(\sin x-2\right)^2\le9\)

=>\(5\le\left(\sin x-2\right)^2+4\le13\)

=>5<=y<=13

=>Tập giá trị là T=[5;13]

\(y_{\min}=5\) khi sin x-2=-1

=>sin x=1

=>\(x=\frac{\pi}{2}+k2\pi\)

\(y_{\max}\) =13 khi sin x-2=-3

=>sin x=-1

=>\(x=-\frac{\pi}{2}+k2\pi\)

a)      

\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) =  - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} =  - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi  + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x =  - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

b)     \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

c)       

\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x =  - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x =  - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x =  - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)

d)      

\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x =  - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x =  - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x =  - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)

e)      

\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)

f)       

\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x =  - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)

a: ĐKXĐ: \(1-\sin\left(x-\frac{\pi}{8}\right)>0\) và \(2x-\frac{\pi}{4}<>\frac{\pi}{2}+k\pi\)

=>\(\sin\left(x-\frac{\pi}{8}\right)<1\) và \(2x<>\frac34\pi+k\pi\)

=>\(\sin\left(x-\frac{\pi}{8}\right)<>1\) và \(x<>\frac38\pi+k\pi\)

=>\(x-\frac{\pi}{8}<>\frac{\pi}{2}+k2\pi\) và \(x<>\frac38\pi+k\pi\)

=>\(x<>\frac58\pi+k2\pi\) và \(x<>\frac38\pi+k\pi\)

=>TXĐ là D=R\{\(\frac58\pi+k2\pi;\frac38\pi+k\pi\) }

b: ĐKXĐ: \(\begin{cases}1-cos\left(x+\frac{\pi}{3}\right)<>0\\ x-\frac{\pi}{4}<>\frac{\pi}{2}+k\pi\end{cases}\Rightarrow\begin{cases}cos\left(x+\frac{\pi}{3}\right)<>1\\ x<>\frac34\pi+k\pi\end{cases}\)

=>\(\begin{cases}x+\frac{\pi}{3}<>k2\pi\\ x<>\frac34\pi+k\pi\end{cases}\Rightarrow\begin{cases}x<>-\frac{\pi}{3}+k2\pi\\ x<>\frac34\pi+k\pi\end{cases}\)

=>TXĐ là D=R\{\(-\frac{\pi}{3}+k2\pi;\frac34\pi+k\pi\) }

c: ĐKXĐ: cosx-cos3x<>0

=>cos3x<>cosx

=>\(\begin{cases}3x<>x+k2\pi\\ 3x<>-x+k2\pi\end{cases}\Rightarrow\begin{cases}2x<>k2\pi\\ 4x<>k2\pi\end{cases}\Rightarrow\begin{cases}x<>k\pi\\ x<>\frac{k\pi}{2}\end{cases}\)

=>\(x<>\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(\frac{k\pi}{2}\) }

d: ĐKXĐ: \(\sin^2x-cos^2x<>0\)

=>\(cos^2x-\sin^2x<>0\)

=>cos 2x<>0

=>\(2x<>\frac{\pi}{2}+k\pi\)

=>\(x<>\frac{\pi}{4}+\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(\frac{\pi}{4}+\frac{k\pi}{2}\) }
e: ĐKXĐ: \(\begin{cases}x+\frac{\pi}{3}<>k\pi\\ 3x-\frac{\pi}{4}<>\frac{\pi}{2}+k\pi\\ 3x-\frac{\pi}{4}<>k\pi\end{cases}\)

=>\(\begin{cases}x<>-\frac{\pi}{3}+k\pi\\ 3x<>\frac34\pi+k\pi\\ 3x<>\frac{\pi}{4}+k\pi\end{cases}\Rightarrow\begin{cases}x<>-\frac{\pi}{3}+k\pi\\ x<>\frac14\pi+\frac{k\pi}{3}\\ x<>\frac{1}{12}\pi+\frac{k\pi}{3}\end{cases}\)

=>TXĐ là D=R\{\(-\frac{\pi}{3}+k\pi;\frac14\pi+\frac{k\pi}{3};\frac{1}{12}\pi+\frac{k\pi}{3}\) }