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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: \(x=\frac{\pi}{6};x=\frac{5\pi}{6}\)
Câu 2 bạn coi lại đề
3.
\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)
\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)
\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
4.
Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm
5.
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)
\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
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.
\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow cos^3\left(x-\frac{\pi}{3}\right)=\frac{1}{8}\)
\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=k2\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{6}-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{\pi}{6}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{6}+\frac{k2\pi}{3}\)
b/
\(\Rightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow-cos2x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos2x=-sin\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{5\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=x+\frac{5\pi}{6}+k2\pi\\2x=-x-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}+k2\pi\\x=-\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
\(\Leftrightarrow1-\frac{1}{2}sin^22x+cos\left(x-\frac{\pi}{4}\right)sin\left(3x-\frac{\pi}{4}\right)-\frac{3}{2}=0\)
Đặt \(x-\frac{\pi}{4}=a\Rightarrow x=a+\frac{\pi}{4}\)
\(\Rightarrow1-\frac{1}{2}sin^2\left(2a+\frac{\pi}{2}\right)+cosa.sin\left(3a+\frac{3\pi}{4}-\frac{\pi}{4}\right)-\frac{3}{2}=0\)
\(\Leftrightarrow1-\frac{1}{2}cos^22a+cosa.cos3a-\frac{3}{2}=0\)
\(\Leftrightarrow2-cos^22a+cos4a+cos2a-3=0\)
\(\Leftrightarrow-cos^22a+2cos^22a-1+cos2a-1=0\)
\(\Leftrightarrow cos^22a+cos2a-2=0\)
\(\Leftrightarrow cos2a=1\Leftrightarrow cos\left(2x-\frac{\pi}{2}\right)=1\)
\(\Leftrightarrow sin2x=1\Rightarrow x=\frac{\pi}{4}+k\pi\)
a/
Đặt \(cosx=t\Rightarrow0< t\le1\)
\(\Rightarrow t^2-2mt+4\left(m-1\right)=0\)
\(\Leftrightarrow t^2-4-2m\left(t-2\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+2-2m\right)=0\)
\(\Leftrightarrow t=2m-2\)
\(\Rightarrow0< 2m-2\le1\Rightarrow1< m\le\frac{3}{2}\)
b.
\(x\in\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\Rightarrow\frac{x}{2}\in\left(-\frac{\pi}{4};\frac{\pi}{4}\right)\)
Đặt \(sin\frac{x}{2}=t\Rightarrow-\frac{\sqrt{2}}{2}< t< \frac{\sqrt{2}}{2}\)
\(\Rightarrow4t^2+2t+m-2=0\Leftrightarrow4t^2+2t-2=-m\)
Xét \(f\left(t\right)=4t^2+2t-2\) trên \(\left(-\frac{\sqrt{2}}{2};\frac{\sqrt{2}}{2}\right)\)
\(f\left(-\frac{\sqrt{2}}{2}\right)=-\sqrt{2}\) ; \(f\left(\frac{\sqrt{2}}{2}\right)=\sqrt{2}\) ; \(f\left(-\frac{1}{4}\right)=-\frac{9}{4}\)
\(\Rightarrow-\frac{9}{4}\le f\left(t\right)< \sqrt{2}\Rightarrow-\frac{9}{4}\le-m< \sqrt{2}\)
\(\Rightarrow-\sqrt{2}< m\le\frac{9}{4}\)