Question

GPT

a) \(sin2x-cosx=0\)

b) \(\left(2cos\frac{x}{2}-1\right)\left(sin\frac{x}{2}+2\right)=0\)

c) \(\left(\sqrt{3}tanx+1\right)\left(sin^2x+1\right)=0\)

d) \(sinx.cosx.cos2x=0\)

e) \(8cos2x.sin2x.cos4x=-\sqrt{2}\)

Answer 1

a.

\(\Leftrightarrow sin2x=cosx\)

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

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

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

b.

\(\Leftrightarrow\left[{}\begin{matrix}cos\frac{x}{2}=\frac{1}{2}\\sin\frac{x}{2}=-2< -1\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k4\pi\\x=-\frac{2\pi}{3}+k4\pi\end{matrix}\right.\)

Answer 2

c.

\(\Leftrightarrow tanx=-\frac{1}{\sqrt{3}}\)

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

d.

\(\Leftrightarrow\frac{1}{2}sin2x.cos2x=0\)

\(\Leftrightarrow\frac{1}{4}sin4x=0\)

\(\Leftrightarrow sin4x=0\)

\(\Leftrightarrow x=\frac{k\pi}{4}\)

e.

\(\Leftrightarrow4sin4x.cos4x=-\sqrt{2}\)

\(\Leftrightarrow2sin8x=-\sqrt{2}\)

\(\Leftrightarrow sin8x=-\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}8x=-\frac{\pi}{4}+k2\pi\\8x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{32}+\frac{k\pi}{4}\\x=\frac{5\pi}{32}+\frac{k\pi}{4}\end{matrix}\right.\)