Bài 1: Hàm số lượng giác

1. \(D=R\)

2. \(sinx\ne0\Leftrightarrow x\ne k\pi\Rightarrow D=R\backslash\left\{k\pi|k\in R\right\}\)

3. \(cos2x\ne0\Leftrightarrow2x\ne\dfrac{\pi}{2}+k\pi\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\Rightarrow D=R\backslash\left\{\dfrac{\pi}{4}+\dfrac{k\pi}{2}|k\in R\right\}\)

4. \(cos\left(x+\dfrac{\pi}{4}\right)\ne0\Leftrightarrow x+\dfrac{\pi}{4}\ne\dfrac{\pi}{2}+k\pi\Leftrightarrow x\ne\dfrac{\pi}{4}+k\pi\Rightarrow D=R\backslash\left\{\dfrac{\pi}{4}+k\pi|k\in R\right\}\)

Do \(\left\{{}\begin{matrix}\left|sinx\right|\le1\\\left|cosx\right|\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}sin^9x\le sin^2x\\cos^{12}x\le cos^2x\end{matrix}\right.\)

\(\Rightarrow sin^9x+cos^{12}x\le sin^2x+cos^2x=1\)

\(y_{max}=1\) khi \(\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

Hàm số xác định  \(\Leftrightarrow\left\{{}\begin{matrix}sinx\ne0\\cos2x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sinx\ne0\\tan^2x\ne1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\dfrac{\pi}{4}+k\pi\\x\ne-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

  chép đề cũng sai

Có: `y=a sinx +b cosx`

`=> -\sqrt(a^2+b^2) <= y <= \sqrt(a^2+b^2)`

`y=sin^4x + cos^4 x+4`

`=(sin^2x)^2 + (cos^2x)^2+4`

`=(sin^2x + 2.sin^2x . cos^2x + cos^2x) - 2sin^2xcos^2x +4`

`= (sin^2x+cos^2x)^2 - 1/2 (2sinxcox).(2sinxcosx) +4`

`= 1^2 -1/2 sin^2 2x +4`

\(I\left(\dfrac{1}{2};\dfrac{3}{4}\right)\)

Nhìn BBT ta thấy \(y_{max}=3\) còn \(y_{min}=\dfrac{3}{4}\)

Dùng CT: \(sin\left(a+b\right)=sina.cosb+cosa.sinb\)

\(y=\sqrt{3}sin2x-cos2x\)

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

\(=2\left(cos\dfrac{\pi}{6}.sin2x-sin\dfrac{\pi}{6}.cos2x\right)\)

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

\(\dfrac{1}{2}sin6x\ne0\)\(\Leftrightarrow sin6x\ne0\) \(\Leftrightarrow6x\ne k\pi\)\(\Leftrightarrow x\ne\dfrac{k\pi}{6}\)

\(\dfrac{1}{2}\ne0\) rồi nên chỉ cần \(sin6x\ne0\)

Tất cả k dưới đây đều là \(k\in Z\)

1a.

\(sin\left(x+2\right)=\dfrac{1}{3}\)

\(\Rightarrow\left[{}\begin{matrix}x+2=arcsin\left(\dfrac{1}{3}\right)+k2\pi\\x+2=\pi-arcsin\left(\dfrac{1}{3}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2+arcsin\left(\dfrac{1}{3}\right)+k2\pi\\x=-2+\pi-arcsin\left(\dfrac{1}{3}\right)+k2\pi\end{matrix}\right.\)

b.

\(sin3x=1\)

\(\Leftrightarrow sin3x=sin\left(\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow3x=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)

1c.

\(sin\left(\dfrac{2x}{3}-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow\dfrac{2x}{3}-\dfrac{\pi}{3}=k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+\dfrac{k3\pi}{2}\)

d.

\(sin\left(2x+20^0\right)=-\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+20^0=-60^0+k360^0\\2x+20^0=240^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-40^0+k180^0\\x=110^0+k180^0\end{matrix}\right.\)

2.

\(\Leftrightarrow sin3x=sinx\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

3.a

\(\Leftrightarrow\left[{}\begin{matrix}x-1=arccos\left(\dfrac{2}{3}\right)+k2\pi\\x-1=-arccos\left(\dfrac{2}{3}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1+arccos\left(\dfrac{2}{3}\right)+k2\pi\\x=1-arccos\left(\dfrac{2}{3}\right)+k2\pi\end{matrix}\right.\)