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c.
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos\left(8x+\frac{2\pi}{3}\right)=\frac{1}{2}-\frac{1}{2}cos\left(\frac{14\pi}{5}-2x\right)\)
\(\Leftrightarrow cos\left(8x+\frac{2\pi}{3}\right)=cos\left(2\pi+\frac{4\pi}{5}-2x\right)\)
\(\Leftrightarrow cos\left(8x+\frac{2\pi}{3}\right)=cos\left(\frac{4\pi}{5}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}8x+\frac{2\pi}{3}=\frac{4\pi}{5}-2x+k2\pi\\8x+\frac{2\pi}{3}=2x-\frac{4\pi}{5}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{75}+\frac{k\pi}{5}\\x=-\frac{11\pi}{45}+\frac{k\pi}{3}\end{matrix}\right.\)
a.
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos4x=\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{2\pi}{3}\right)\)
\(\Leftrightarrow cos4x=-cos\left(2x+\frac{2\pi}{3}\right)\)
\(\Leftrightarrow cos4x=cos\left(\frac{\pi}{3}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{3}-2x+k2\pi\\4x=2x-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{18}+\frac{k\pi}{3}\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
b.
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos\left(10x+\frac{2\pi}{3}\right)-\frac{1}{2}-\frac{1}{2}cos\left(6x+\frac{\pi}{2}\right)=0\)
\(\Leftrightarrow cos\left(10x+\frac{2\pi}{3}\right)=-cos\left(6x+\frac{\pi}{2}\right)\)
\(\Leftrightarrow cos\left(10x+\frac{2\pi}{3}\right)=cos\left(\frac{\pi}{2}-6x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}10x+\frac{2\pi}{3}=\frac{\pi}{2}-6x+k2\pi\\10x+\frac{2\pi}{3}=6x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{96}+\frac{k\pi}{8}\\x=-\frac{7\pi}{24}+\frac{k\pi}{2}\end{matrix}\right.\)
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):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
7.
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow8cosx=\frac{\sqrt{3}cosx+sinx}{sinx.cosx}\)
\(\Leftrightarrow8cosx.sinx.cosx=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow4sin2x.cosx=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow2sin3x+2sinx=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow2sin3x=\sqrt{3}cosx-sinx\)
\(\Leftrightarrow sin3x=\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx\)
\(\Leftrightarrow sin\left(-3x\right)=sin\left(x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x=x-\frac{\pi}{3}+k2\pi\\-3x=\frac{4\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{2}\\x=-\frac{2\pi}{3}+k\pi\end{matrix}\right.\)
5.
\(sin\left(2x+\frac{\pi}{2}+2\pi\right)-2cos\left(x+\frac{\pi}{2}-4\pi\right)=1+2sinx\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}\right)-2cos\left(x+\frac{\pi}{2}\right)=1+2sinx\)
\(\Leftrightarrow cos2x+2sinx=1+2sinx\)
\(\Leftrightarrow cos2x=1\)
\(\Rightarrow x=k\pi\)
6.
\(sin^22x-cos^28x=sin\left(10x+\frac{\pi}{2}+8\pi\right)\)
\(\Leftrightarrow\frac{1-cos4x}{2}-\frac{1+cos16x}{2}=sin\left(10x+\frac{\pi}{2}\right)\)
\(\Leftrightarrow-\left(cos4x+cos16x\right)=2cos10x\)
\(\Leftrightarrow-2cos10x.cos6x=2cos10x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos10x=0\\cos6x=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}10x=\frac{\pi}{2}+k\pi\\6x=\pi+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{20}+\frac{k\pi}{10}\\x=\frac{\pi}{6}+\frac{k\pi}{3}\end{matrix}\right.\)
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Chứng minh các biểu thức đã cho không phụ thuộc vào x.
Từ đó suy ra f'(x)=0
a) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
b) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
c) f(x)=\(\frac{1}{4}\)(\(\sqrt{2}\)-\(\sqrt{6}\))=>f'(x)=0
d,f(x)=\(\frac{3}{2}\)=>f'(x)=0
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.
\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
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
\(y_{min}=y\left(\frac{3}{2}\right)=cos\left(\frac{3}{2}\right)\)
\(y_{max}=y\left(\frac{1}{4}\right)=cos\left(\frac{1}{4}\right)\)
\(A=2\cdot cos\left(\dfrac{\Omega}{2}+x\right)+sin\left(5\Omega-x\right)+sin\left(\dfrac{3\Omega}{2}+x\right)+cos\left(\dfrac{\Omega}{2}+x\right)\)
\(=3\cdot cos\left(\dfrac{\Omega}{2}+x\right)+sin\left(\Omega-x\right)+sin\left(\dfrac{\Omega}{2}+\Omega+x\right)\)
\(=-3\cdot sinx+sinx+cos\left(\Omega+x\right)\)
\(=-2\cdot sinx-cosx\)
\(B=sin\left(\Omega+x\right)-cos\left(\dfrac{\Omega}{2}+x\right)+cot\left(2\Omega-x\right)+tan\left(\dfrac{2\Omega}{2}+x\right)\)
\(=-sinx+sinx+cot\left(-x\right)+tan\left(x\right)\)
\(=tanx-cotx=tanx-\dfrac{1}{tanx}=\dfrac{tan^2x-1}{tanx}\)