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a/
\(\Leftrightarrow sin\left(x+\frac{\pi}{8}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{8}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{8}=\frac{5\pi}{6}+l2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}+k2\pi\\x=\frac{17\pi}{24}+l2\pi\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-\pi\le\frac{\pi}{24}+k2\pi\le\pi\\-\pi\le\frac{17\pi}{24}+l2\pi\le\pi\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=0\\l=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}\\x=\frac{17\pi}{24}\end{matrix}\right.\) \(\Rightarrow\sum x=\frac{\pi}{24}+\frac{17\pi}{24}=\frac{3\pi}{4}\)
2.
\(4sin^22x-1=0\Leftrightarrow2-2cos4x-1=0\)
\(\Leftrightarrow cos4x=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+\frac{l\pi}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-\frac{\pi}{2}\le\frac{\pi}{12}+\frac{k\pi}{2}\le\frac{\pi}{2}\\-\frac{\pi}{2}\le-\frac{\pi}{12}+\frac{l\pi}{2}\le\frac{\pi}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=\left\{-1;0\right\}\\l=\left\{0;1\right\}\end{matrix}\right.\)
\(\Rightarrow x=\left\{-\frac{5\pi}{12};\frac{\pi}{12};-\frac{\pi}{12};\frac{5\pi}{12}\right\}\Rightarrow\sum x=0\)
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: \(x=\frac{\pi}{6};x=\frac{5\pi}{6}\)
