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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
\(cos\cdot\left(3x-\dfrac{\pi}{6}\right)=sin\cdot\left(x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow cos\cdot\left(3x-\dfrac{\pi}{6}\right)=cos\cdot\left(\dfrac{\pi}{2}-x-\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow cos\cdot\left(3x-\dfrac{\pi}{6}\right)=cos\cdot\left(\dfrac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{6}=\dfrac{\pi}{4}-x+k2\pi\\3x-\dfrac{\pi}{6}=\dfrac{-\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{5\pi}{12}+k2\pi\\2x=\dfrac{-\pi}{12}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{48}+\dfrac{k\pi}{2}\\x=\dfrac{-\pi}{24}+k\pi\end{matrix}\right.\left(k\in Z\right)\)
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.
\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)