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1.
Từ đường tròn lượng giác ta thấy pt đã cho có nghiệm duy nhất thuộc \(\left[-\frac{\pi}{2};\frac{\pi}{3}\right]\) khi và chỉ khi:
\(\left[{}\begin{matrix}2m=1\\0\le2m< \frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m=\frac{1}{2}\\0\le m< \frac{1}{4}\end{matrix}\right.\)
2.
\(\Leftrightarrow3x-\frac{\pi}{3}=x+\frac{\pi}{4}+k\pi\)
\(\Leftrightarrow x=\frac{7\pi}{24}+\frac{k\pi}{2}\)
\(-\pi< \frac{7\pi}{24}+\frac{k\pi}{2}< \pi\Rightarrow-\frac{31}{12}< k< \frac{17}{12}\)
\(\Rightarrow k=\left\{-2;-1;0;1\right\}\) có 4 nghiệm
3.
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\2x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\) có 4 điểm biểu diễn
1.
\(\Leftrightarrow2x-\frac{\pi}{4}=x+\frac{\pi}{3}+k\pi\)
\(\Rightarrow x=\frac{7\pi}{12}+k\pi\)
\(-\pi< \frac{7\pi}{12}+k\pi< \pi\Rightarrow-\frac{19}{12}< k< \frac{5}{12}\Rightarrow k=\left\{-1;0\right\}\) có 2 nghiệm
\(x=\left\{-\frac{5\pi}{12};\frac{7\pi}{12}\right\}\)
2.
\(\Leftrightarrow3x-\frac{\pi}{3}=\frac{\pi}{2}+k\pi\)
\(\Rightarrow x=\frac{5\pi}{18}+\frac{k\pi}{3}\)
Nghiệm âm lớn nhất là \(x=-\frac{\pi}{18}\) khi \(k=-1\)
3.
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{3\pi}{4}=\frac{\pi}{3}+k2\pi\\x-\frac{3\pi}{4}=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{13\pi}{12}+k2\pi\\x=\frac{17\pi}{12}+k2\pi\end{matrix}\right.\)
Nghiệm âm lớn nhất \(x=-\frac{7\pi}{12}\) ; nghiệm dương nhỏ nhất \(x=\frac{13\pi}{12}\)
Tổng nghiệm: \(\frac{\pi}{2}\)
Câu 1 với câu 2 sai đề, sin và cos nằm trong [-1;1], mà căn 2 với căn 3 lớn hơn 1 rồi
3/ \(\sin x=\cos2x=\sin\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}-2x+k2\pi\\x=\pi-\frac{\pi}{2}+2x+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\frac{2}{3}\pi\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
4/ \(\Leftrightarrow\cos^2x-2\sin x\cos x=0\)
Xét \(\cos x=0\) là nghiệm của pt \(\Rightarrow x=\frac{\pi}{2}+k\pi\)
\(\cos x\ne0\Rightarrow1-2\tan x=0\Leftrightarrow\tan x=\frac{1}{2}\Rightarrow x=...\)
5/ \(\Leftrightarrow\sin\left(2x+1\right)=-\cos\left(3x-1\right)=\cos\left(\pi-3x+1\right)=\sin\left(\frac{\pi}{2}-\pi+3x-1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\frac{\pi}{2}-\pi+3x-1\\2x+1=\pi-\frac{\pi}{2}+\pi-3x+1\end{matrix}\right.\Leftrightarrow....\)
6/ \(\Leftrightarrow\cos\left(\pi\left(x-\frac{1}{3}\right)\right)=\frac{1}{2}\Leftrightarrow\pi\left(x-\frac{1}{3}\right)=\pm\frac{\pi}{3}+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{1}{3}+2k\Rightarrow x=\frac{2}{3}+2k\left(1\right)\\x-\frac{1}{3}=-\frac{1}{3}+2k\Rightarrow x=2k\left(2\right)\end{matrix}\right.\)
\(\left(1\right):-\pi< x< \pi\Rightarrow-\pi< \frac{2}{3}+2k< \pi\) (Ủa đề bài sai hay sao ý nhỉ?)
7/ \(\Leftrightarrow\left[{}\begin{matrix}5x+\frac{\pi}{3}=\frac{\pi}{2}-2x+\frac{\pi}{3}\\5x+\frac{\pi}{3}=\pi-\frac{\pi}{2}+2x-\frac{\pi}{3}\end{matrix}\right.\Leftrightarrow...\)
Thui, để đây bao giờ...hết lười thì làm tiếp :(
7)
\(sin\left(5x+\frac{\pi}{3}\right)=cos\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(5x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{2}-2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+\frac{\pi}{3}=\frac{\pi}{2}-2x-\frac{\pi}{3}+k2\pi\\5x+\frac{\pi}{3}=\pi-\left(\frac{\pi}{2}-2x-\frac{\pi}{3}\right)+k2\pi\end{matrix}\right.\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{42}+k\frac{2\pi}{7}\\x=\frac{\pi}{6}+k\frac{2\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)
Do:\(0< x< \pi\)
\(Với:x=\frac{-\pi}{42}+k\frac{2\pi}{7}\left(k\in Z\right)\Rightarrow khôngtìmđượck\)
\(Với:x=\frac{\pi}{6}+k\frac{2\pi}{3}\left(k\in Z\right)\Leftrightarrow\frac{1}{4}< k< \frac{5}{4}\Rightarrow k=\left\{0;1\right\}\Rightarrow\left[{}\begin{matrix}k=0\Rightarrow x=\frac{\pi}{6}\\k=1\Rightarrow x=\frac{5\pi}{6}\end{matrix}\right.\)
Vậy nghiệm của pt là: \(x=\frac{\pi}{6};x=\frac{5\pi}{6}\)
\(sin3x=-\frac{\sqrt{3}}{2}=sin\left(-\frac{\pi}{3}\right)\)
\(\Rightarrow\left[{}\begin{matrix}3x=-\frac{\pi}{3}+k2\pi\\3x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{9}+\frac{k2\pi}{3}\\x=\frac{4\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)
\(sin\left(2x-\frac{\pi}{7}\right)=\frac{\sqrt{2}}{2}=sin\left(\frac{\pi}{4}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x-\frac{\pi}{7}=\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{7}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{11\pi}{56}+k\pi\\x=\frac{25\pi}{56}+k\pi\end{matrix}\right.\)
\(sin\left(4x+1\right)=\frac{3}{5}=sina\) (với góc a sao cho \(sina=\frac{3}{5}\))
\(\Rightarrow\left[{}\begin{matrix}4x+1=a+k2\pi\\4x+1=\pi-a+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{a}{4}-\frac{1}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{4}-\frac{a}{4}-\frac{1}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
\(sin\left(2x+\frac{\pi}{7}\right)=sin\left(x-\frac{3\pi}{7}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x+\frac{\pi}{7}=x-\frac{3\pi}{7}+k2\pi\\2x+\frac{\pi}{7}=\pi-x+\frac{3\pi}{7}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{4\pi}{7}+k2\pi\\x=\frac{3\pi}{7}+\frac{k2\pi}{3}\end{matrix}\right.\)
\(sin\left(4x+\frac{\pi}{7}\right)=\frac{1}{4}\)
Đặt \(\frac{1}{4}=sina\Rightarrow sin\left(4x+\frac{\pi}{7}\right)=sina\)
\(\Rightarrow\left[{}\begin{matrix}4x+\frac{\pi}{7}=a+k2\pi\\4x+\frac{\pi}{7}=\pi-a+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{28}+\frac{a}{4}+\frac{k\pi}{2}\\x=\frac{3\pi}{14}-\frac{a}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
c.
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=-sin\left(x-\frac{2\pi}{5}-\pi\right)\)
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=sin\left(x-\frac{2\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{2\pi}{3}=x-\frac{2\pi}{5}+k2\pi\\3x+\frac{2\pi}{3}=\frac{7\pi}{5}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{8\pi}{15}+k\pi\\x=\frac{11\pi}{60}+\frac{k\pi}{2}\end{matrix}\right.\)
d.
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=\frac{\pi}{4}+x+k2\pi\\4x+\frac{\pi}{3}=-\frac{\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)
a.
\(sin\left(2x+1\right)=-cos\left(3x-1\right)\)
\(\Leftrightarrow sin\left(2x+1\right)=sin\left(3x-1-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1-\frac{\pi}{2}=2x+1+k2\pi\\3x-1-\frac{\pi}{2}=\pi-2x-1+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+2+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
b.
\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)