Question

Answer 1

21.

ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(\sqrt{3}+3tanx=0\)

\(\Leftrightarrow tanx=-\dfrac{\sqrt{3}}{3}\)

\(\Leftrightarrow x=-\dfrac{\pi}{6}+k\pi\)

Answer 2

22.

\(sin3x=sinx\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=x+k2\pi\\3x=\pi-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Answer 3

23.

\(sinx.cosx=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow2x=k\pi\)

\(\Leftrightarrow x=\dfrac{k\pi}{2}\)

Answer 4

24.

\(cos3x=cosx\)

\(\Leftrightarrow3x=\pm x+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=k2\pi\\4x=k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

Answer 5

25.

\(sin3x=cosx\)

\(\Leftrightarrow cos\left(\dfrac{\pi}{2}-3x\right)=cosx\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)