Question

giải các pt (nhân tử chung)

a) \(sin4x=2cos2x.cosx\)

b) \(\left(cosx+sin2x\right).sin2x=0\)

c) \(cosx+cos2x+cos3x+cos4x=0\)

d) \(sin3x-sinx+sin2x=0\)

Answer 1

\(sin4x-2cos2x.cosx=0\)

\(\Leftrightarrow2sin2x.cos2x-2cos2x.cosx=0\)

\(\Leftrightarrow cos2x\left(sin2x-cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\\sin2x-cosx=0\left(1\right)\end{matrix}\right.\)

Xét (1)

\(\Leftrightarrow sin2x=cosx=sin\left(\frac{\pi}{2}-x\right)\)

\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-x+k2\pi\\2x=x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

Answer 2

\(\left(cosx+sin2x\right).sin2x=0\)

\(\Leftrightarrow\left(cosx+2sinx.cosx\right).2sinx.cosx=0\)

\(\Leftrightarrow\left(1+2sinx\right)sinx.cos^2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}1+2sinx=0\\sinx.cosx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin2x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\2x=k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\end{matrix}\right.\)

Answer 3

\(cosx+cos3x+cos2x+cos4x=0\)

\(\Leftrightarrow2cos2x.cosx+2cos3x.cosx=0\)

\(\Leftrightarrow cosx\left(cos2x+cos3x\right)=0\)

\(\Leftrightarrow2cosx.cos\frac{5x}{2}.cos\frac{x}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos\frac{x}{2}=0\\cos\frac{5x}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\end{matrix}\right.\)

Answer 4

\(sin3x-sinx+sin2x=0\)

\(\Leftrightarrow2cos2x.sinx+2sinx.cosx=0\)

\(\Leftrightarrow sinx\left(cos2x+cosx\right)=0\)

\(\Leftrightarrow2sinx.cos\frac{3x}{2}.cos\frac{x}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=0\\cos\frac{x}{2}=0\\cos\frac{3x}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{3x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\pi+k2\pi\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)