Question

GPT

a) \(\left(2sinx-1\right)\left(\sqrt{3}cosx-5\right)=0\)

b) \(sin2x.cos2x.cos4x+\frac{1}{8}=0\)

c) \(sin4x+\sqrt{3}sin2x=0\)

d) \(\left(\sqrt{2}sin2x+2\right)\left(2cosx+\sqrt{2}\right)=0\)

Answer 1

a.

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cosx=\frac{5}{\sqrt{3}}>1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

b.

\(\Leftrightarrow\frac{1}{2}sin4x.cos4x+\frac{1}{8}=0\)

\(\Leftrightarrow\frac{1}{4}sin8x+\frac{1}{8}=0\)

\(\Leftrightarrow sin8x=-\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}8x=-\frac{\pi}{6}+k2\pi\\8x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{48}+\frac{k\pi}{4}\\x=\frac{7\pi}{48}+\frac{k\pi}{4}\end{matrix}\right.\)

Answer 2

c.

\(\Leftrightarrow2sin2x.cos2x+\sqrt{3}sin2x=0\)

\(\Leftrightarrow sin2x\left(2cos2x+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\frac{5\pi}{6}+k2\pi\\2x=-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{5\pi}{12}+k\pi\\x=-\frac{5\pi}{12}+k\pi\end{matrix}\right.\)

d.

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-\sqrt{2}< -1\left(l\right)\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{\pi}{4}+k2\pi\\2x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

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