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
\(\Leftrightarrow\left[{}\begin{matrix}x-1=arccos\left(\dfrac{2}{3}\right)+k2\pi\\x-1=-arccos\left(\dfrac{2}{3}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1+arccos\left(\dfrac{2}{3}\right)+k2\pi\\x=1-arccos\left(\dfrac{2}{3}\right)+k2\pi\end{matrix}\right.\)
3b.
\(\Leftrightarrow\left[{}\begin{matrix}3x=12^0+k360^0\\3x=-12^0+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4^0+k120^0\\x=-4^0+k120^0\end{matrix}\right.\)
3c.
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x}{2}-\dfrac{\pi}{4}=\dfrac{2\pi}{3}+k2\pi\\\dfrac{3x}{2}-\dfrac{\pi}{4}=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x}{2}=\dfrac{11\pi}{12}+k2\pi\\\dfrac{3x}{2}=-\dfrac{5\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11\pi}{18}+\dfrac{k4\pi}{3}\\x=-\dfrac{5\pi}{18}+\dfrac{k4\pi}{3}\end{matrix}\right.\)
3d.
\(\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{2}cos4x=\dfrac{1}{4}\)
\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=\dfrac{2\pi}{3}+k2\pi\\cos4x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\end{matrix}\right.\)
4.
ĐKXĐ: \(sin2x\ne1\Leftrightarrow x\ne\dfrac{\pi}{4}+k\pi\)
Pt tương đương:
\(2cos2x=0\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
Kết hợp ĐKXĐ ta được: \(x=-\dfrac{\pi}{4}+k\pi\)
5a/
\(\Leftrightarrow x-15^0=30^0+k180^0\)
\(\Leftrightarrow x=45^0+k180^0\)
b.
\(\Leftrightarrow3x-1=-\dfrac{\pi}{6}+k\pi\)
\(\Leftrightarrow3x=1-\dfrac{\pi}{6}+k\pi\)
\(\Leftrightarrow x=\dfrac{1}{3}-\dfrac{\pi}{18}+\dfrac{k\pi}{3}\)
c.
ĐKXĐ: \(cosx\ne0\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\)
Pt tương đương: \(\left[{}\begin{matrix}cos2x=0\\tanx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\x=k\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=k\pi\end{matrix}\right.\)
5d.
ĐKXĐ: \(sinx\ne0\Rightarrow x\ne k\pi\)
Pt tương đương:
\(\left[{}\begin{matrix}sin3x=0\\cotx=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x=k\pi\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{3}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
Kết hợp ĐKXĐ ta được: \(\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\\x=-\dfrac{\pi}{3}+k\pi\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
6.
Hai hàm số có giá trị bằng nhau khi:
\(tan2x=tan\left(\dfrac{\pi}{4}-x\right)\)
\(\Leftrightarrow2x=\dfrac{\pi}{4}-x+k\pi\)
\(\Leftrightarrow3x=\dfrac{\pi}{4}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\)
7a.
\(\Leftrightarrow cos5x=sin3x\)
\(\Leftrightarrow cos5x=cos\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{2}-3x+k2\pi\\5x=3x-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}8x=\dfrac{\pi}{2}+k2\pi\\2x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{4}\\x=-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)
b.
ĐKXĐ: ...
\(tan3x=\dfrac{1}{tanx}=cotx\)
\(\Leftrightarrow tan3x=tan\left(\dfrac{\pi}{2}-x\right)\)
\(\Leftrightarrow3x=\dfrac{\pi}{2}-x+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)
7.
a, \(sin3x-cos5x=0\)
\(\Leftrightarrow sin3x=cos5x\)
\(\Leftrightarrow sin3x=sin\left(\dfrac{\pi}{2}-5x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{2}-5x+k2\pi\\3x=\pi-\dfrac{\pi}{2}+5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{4}\\x=-\dfrac{\pi}{4}-k\pi\end{matrix}\right.\)
Vậy phương trình có nghiệm \(x=\dfrac{\pi}{16}+\dfrac{k\pi}{4};x=-\dfrac{\pi}{4}-k\pi\)
b, ĐK: \(x\ne\dfrac{\pi}{6}+\dfrac{k\pi}{3}\)
\(tan3x.tanx=1\)
\(\Leftrightarrow\dfrac{sin3x.sinx}{cos3x.cosx}=1\)
\(\Leftrightarrow sin3x.sinx=cos3x.cosx\)
\(\Leftrightarrow\dfrac{1}{2}\left(cos2x-cos4x\right)=\dfrac{1}{2}\left(cos4x+cos2x\right)\)
\(\Leftrightarrow2cos4x=0\)
\(\Leftrightarrow4x=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\left(tm\right)\)
Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)
.Nhưng khoảng từ pi/2 đến -pi/4 thì vẫn bao gồm đáp án A mà sao mình lại chọn ạ
