Question

a) \(\sin2x=\dfrac{-1}{2}\)

b) \(\sin\dfrac{x}{2}=-1\)

c) \(\sin3x=\dfrac{1}{3}\)

d) \(\sin4x=-\sin x\)

Answer 1

\(a,\sin2x=\dfrac{-1}{2}\Leftrightarrow\sin2x=\sin\left(-\dfrac{\pi}{6}\right)\\ \Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{6}+k2\pi\\2x=\pi+\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k2\pi\\x=\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)

Answer 2

a, \(sin2x=-\dfrac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k\pi\\x=\dfrac{7\pi}{12}+k\pi\end{matrix}\right.\)

Answer 3

b, \(sin\dfrac{x}{2}=-1\)

\(\Leftrightarrow\dfrac{x}{2}=-\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=-\pi+k2\pi\)

Answer 4

c, \(sin3x=\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=arrcsin\dfrac{1}{3}+k2\pi\\3x=\pi-arrcsin\dfrac{1}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}arrcsin\dfrac{1}{3}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{3}-\dfrac{1}{3}arrcsin\dfrac{1}{3}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

Answer 5

d, \(sin4x=-sinx\)

\(\Leftrightarrow2sin\dfrac{5x}{2}.cos\dfrac{3x}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\dfrac{5x}{2}=0\\cos\dfrac{3x}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5x}{2}=k\pi\\\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{3}+\dfrac{k2\pi}{3}\end{matrix}\right.\)