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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/
\(\Rightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow-cos2x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos2x=-sin\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{5\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=x+\frac{5\pi}{6}+k2\pi\\2x=-x-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}+k2\pi\\x=-\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
c.
\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=k2\pi\end{matrix}\right.\)
\(\Rightarrow x=\left\{0;\frac{2\pi}{3}\right\}\)
d.
\(\Leftrightarrow cos^2x\left(cosx-2\right)=0\)
\(\Leftrightarrow cosx=0\)
\(\Leftrightarrow x=90^0+k180^0\)
\(\Rightarrow x=\left\{90^0;270^0;450^0;630^0\right\}\)
a.
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=\left\{\frac{4\pi}{3};\frac{5\pi}{3}\right\}\)
b.
\(\Leftrightarrow sin2x=-\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-30^0+k360^0\\2x=210^0+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-15^0+k180^0\\x=105^0+k180^0\end{matrix}\right.\)
Pt vô nghiệm trên khoảng đã cho
a. \(sin\left(4x+\pi\right)=sin35^o\)
\(\Leftrightarrow sin\left(4x+180^o\right)=sin35^o\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+180^o=35^o+k.360^o,k\in Z\\4x+180^o=180^o-35^o+k.360^o,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-145^o+k.360^o,k\in Z\\4x=-35^o+k.360^o,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{145^o}{4}+k.90,k\in Z\\x=-\frac{35^o}{4}+k.90^o,k\in Z\end{matrix}\right.\)
Vậy.....
b.\(sin4x=\frac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=arcsin\left(\frac{1}{5}\right)+k2\pi,k\in Z\\4x=\pi-arcsin\left(\frac{1}{5}\right)+k2\pi,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{arcsin\left(\frac{1}{5}\right)}{4}+\frac{k\pi}{2},k\in Z\\x=\frac{\pi}{4}-\frac{arcsin\left(\frac{1}{5}\right)}{4}+\frac{k\pi}{2},k\in Z\end{matrix}\right.\)
Vậy....
c. \(sin\left(x+\frac{8\pi}{7}\right)=3\)
Ta có: \(-1\le sinx\le1\)
\(\Rightarrow-1\le sin\left(3x+\frac{8\pi}{7}\right)\le1\)
Do đó phương trình trên vô nghiệm
d. \(sinx=-7\)
Ta có: \(-1\le sinx\le1\)
Do đó phương trình trên vô nghiệm
e. \(sin\left(3x+\pi\right)=sin\left(2x-3\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\pi=2x-3\pi+k2\pi,k\in Z\\3x+\pi=\pi-2x+3\pi+k2\pi,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\pi+k2\pi,k\in Z\\5x=3\pi+k2\pi,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\pi+k2\pi,k\in Z\\x=\frac{3}{5}\pi+\frac{k2\pi}{5},k\in Z\end{matrix}\right.\)
Vậy......
f. \(sin\left(4x-\frac{\pi}{2}\right)=sin\left(\pi-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{2}=\pi-2x+k2\pi,k\in Z\\4x-\frac{\pi}{2}=\pi-\pi+2x+k2\pi,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=\frac{3}{2}\pi+k2\pi,k\in Z\\2x=\frac{\pi}{2}+k2\pi,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{3},k\in Z\\x=\frac{\pi}{4}+k\pi,k\in Z\end{matrix}\right.\)
Vậy......
a/ Trên đoạn xét thuộc cung thứ 4, sinx đồng biến
\(\Rightarrow y_{min}=sin\left(-\frac{\pi}{2}\right)=-1\) ; \(y_{max}=sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}\)
b/ Trên đoạn xét thuộc cung phần tư thứ nhất và thứ 4, cosx luôn không âm
\(\Rightarrow y_{min}=cos\left(-\frac{\pi}{2}\right)=cos\left(\frac{\pi}{2}\right)=0\) ; \(y_{max}=cos0=1\)
c/ Trên đoạn xét thuộc cung phần tư thứ tư, sinx đồng biến
\(y_{min}=sin\left(-\frac{\pi}{2}\right)=-1\) ; \(y_{max}=sin0=0\)
d/ Trên đoạn xét thuộc cung phần tư thứ nhất (\(0< \frac{1}{4}< \frac{3}{2}< \frac{\pi}{2}\))
\(\Rightarrow cosx\) nghịch biến
\(y_{min}=y\left(\frac{3}{2}\right)=cos\left(\frac{3}{2}\right)\)
\(y_{max}=y\left(\frac{1}{4}\right)=cos\left(\frac{1}{4}\right)\)
d/
\(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}cos\left(\frac{x}{5}-\frac{\pi}{12}\right)-\frac{\sqrt{3}}{2}sin\left(\frac{x}{5}-\frac{\pi}{12}\right)\right)=sin\left(\frac{x}{5}+\frac{2\pi}{3}\right)-sin\left(\frac{3x}{5}+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\sqrt{2}cos\left(\frac{x}{5}-\frac{\pi}{12}+\frac{\pi}{3}\right)=2cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)sin\left(\frac{\pi}{4}-\frac{x}{5}\right)\)
\(\Leftrightarrow cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=\sqrt{2}cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)cos\left(\frac{x}{5}-\frac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=0\\cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{5}-\frac{\pi}{4}=\frac{\pi}{2}+k\pi\\\frac{2x}{5}+\frac{5\pi}{12}=\frac{\pi}{4}+k2\pi\\\frac{2x}{5}+\frac{5\pi}{12}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{15\pi}{4}+k5\pi\\x=-\frac{5\pi}{12}+k5\pi\\x=-\frac{5\pi}{3}+k5\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\sqrt{3}sin\left(x-\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}-x\right)=2sin1972x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin\left(x-\frac{\pi}{3}\right)+\frac{1}{2}cos\left(x-\frac{\pi}{3}\right)=sin1972x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}+\frac{\pi}{6}\right)=sin1972x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin1972x\)
\(\Leftrightarrow\left[{}\begin{matrix}1972x=x-\frac{\pi}{6}+k2\pi\\1972x=\frac{7\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{11826}+\frac{k2\pi}{1971}\\x=\frac{7\pi}{11838}+\frac{k2\pi}{1973}\end{matrix}\right.\)
Câu 1:
\(\Leftrightarrow sinx.cos\frac{\pi}{3}-cosx.sin\frac{\pi}{3}+2\left(cosx.cos\frac{\pi}{6}+sinx.sin\frac{\pi}{6}\right)=0\)
\(\Leftrightarrow sinx+\frac{1}{\sqrt{3}}cosx=0\)
Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cosx\)
\(tanx+\frac{1}{\sqrt{3}}=0\Rightarrow tanx=-\frac{1}{\sqrt{3}}\Rightarrow x=\frac{\pi}{6}+k\pi\)
Câu 2:
\(\Leftrightarrow1-cos6x=1+cos2x\)
\(\Leftrightarrow-cos6x=cos2x\)
\(\Leftrightarrow cos\left(\pi-6x\right)=cos2x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\pi-6x+k2\pi\\2x=6x-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
Câu 3:
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}-4\pi\right)+cos2x=1\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}\right)+cos2x=1\)
\(\Leftrightarrow cos2x+cos2x=1\)
\(\Leftrightarrow cos2x=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
Câu 4:
\(\sqrt{2}\left(cosx.cos\frac{3\pi}{4}+sinx.sin\frac{3\pi}{4}\right)=1+sinx\)
\(\Leftrightarrow-cosx+sinx=1+sinx\)
\(\Leftrightarrow cosx=-1\Rightarrow x=\pi+k\pi2\)
Câu 5:
Giống câu 3, chắc bạn ghi nhầm đề