1. Hàm số xác định `<=> 1-cosx \ne 0<=>cosx \ne 1<=>x \ne k2π`

Vì: `1+cosx >=0 forallx ; 1-cosx >=0 forall x`

2. Hàm số xác định `<=> sin^2x \ne cos^2x <=> (1-cos2x)/2 \ne (1+cos2x)/2`

`<=>cos2x \ne 0<=> 2x \ne π/2+kπ <=> x \ne π/4+kπ/2`

3. Hàm số xác định `<=> cos2x \ne 0<=> x \ne π/4+kπ/2 (k \in ZZ)`.

\(Vì-1\le\sin x\le1\)

\(\Rightarrow-5\le5\sin x\le5\)

\(\Rightarrow-6\le5\sin x-1\le4\)

Vì \(5\sin x-1\ge0\Leftrightarrow\sin x\ge\dfrac{1}{5}\)

\(\Rightarrow\sqrt{\dfrac{1}{5}}\le\sqrt{5\sin x-1}\le\sqrt{4}\)

\(\Rightarrow\dfrac{1}{\sqrt{5}}\le\sqrt{5\sin x-1}\le2\)

\(\Rightarrow\dfrac{1}{\sqrt{5}}+2\le\sqrt{5\sin x-1}+2\le4\)

\(\Rightarrow\dfrac{1}{\sqrt{5}}+2\le y\le4\)

\(Vậy\) \(y_{max}=4\)

        \(y_{min}=\dfrac{1}{\sqrt{5}}+2\)

[ 1/5 ; 1 )

1/ mình giải ở bài kia rồi

Bạn viết đề rõ ràng hơn đi

VD: 1/y=\(\dfrac{7sin\left(x-\dfrac{\Pi}{5}\right)}{cos\left(x-\Pi\right)}\)

\(\Leftrightarrow y=3\cos2x-2\left(\dfrac{1+\cos2x}{2}\right)+5\)

\(\Leftrightarrow y=3\cos2x-1-\cos2x+5\)

\(\Leftrightarrow y=2\cos2x+4\)

\(Vì\) \(-1\le\cos2x\le1\)

\(\Rightarrow-2\le2\cos2x\le2\)

\(\Rightarrow2\le2\cos2x+4\le6\)

\(\Rightarrow2\le y\le6\)

\(Vậy\) \(y_{max}=6\)

        \(y_{min}=2\)

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 = \(\sqrt{5}\)

3.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{2}-3x+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ

1.

\(sin\left(sinx\right)=0\)

\(\Leftrightarrow sinx=k\pi\) (1)

Do \(-1\le sinx\le1\Rightarrow-1\le k\pi\le1\)

\(\Rightarrow-\dfrac{1}{\pi}\le k\le\dfrac{1}{\pi}\Rightarrow k=0\) do \(k\in Z\)

Thế vào (1)

\(\Rightarrow sinx=0\Rightarrow x=n\pi\)

2.

\(sin\left(cosx\right)=1\Leftrightarrow cosx=\dfrac{\pi}{2}+k2\pi\)

Do \(-1\le cosx\le1\Rightarrow-1\le\dfrac{\pi}{2}+k2\pi\le1\)

\(\Rightarrow-\dfrac{1}{2\pi}-\dfrac{1}{4}\le k\le\dfrac{1}{2\pi}-\dfrac{1}{4}\) 

\(\Rightarrow\) Không tồn tại k thỏa mãn

Pt vô nghiệm

xét hàm số y=\(\sin\left(cos^2x\right)cos\left(sin^2x\right)\) =

\(\frac{sin\left(cos^2x+sin^2x\right)+sin\left(cos^2x-sin^2x\right)}{2}=\frac{sin1+sin\left(cós2x\right)}{2}\)