Question

Giải các phương trình sau

1) \(\cos^5\frac{x}{2}\)\(\sin\frac{x}{2}\) - \(\sin^5\frac{x}{2}\)\(\cos\frac{x}{2}\) = \(\frac{\sqrt{3}}{8}\)

2) 8\(\cos x.\cos\left(\frac{\pi}{3}+x\right).\cos\left(\frac{\pi}{3}-x\right)\) = \(\sqrt{3}\)

3) \(\frac{1}{\cos x}+\frac{1}{\sin2x}=\frac{2}{\sin4x}\)

Answer 1

1.

\(\Leftrightarrow2sin\frac{x}{2}cos\frac{x}{2}\left(cos^4\frac{x}{2}-sin^4\frac{x}{2}\right)=\frac{\sqrt{3}}{4}\)

\(\Leftrightarrow sinx\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\left(cos^2\frac{x}{2}+sin^2\frac{x}{2}\right)=\frac{\sqrt{3}}{4}\)

\(\Leftrightarrow sinx.cosx=\frac{\sqrt{3}}{4}\)

\(\Leftrightarrow\frac{1}{2}sin2x=\frac{\sqrt{3}}{4}\)

\(\Leftrightarrow sin2x=\frac{\sqrt{3}}{2}\)

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

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

Answer 2

2.

\(\Leftrightarrow4cosx\left(cos\frac{2\pi}{3}+cos2x\right)=\sqrt{3}\)

\(\Leftrightarrow4cosx\left(cos2x-\frac{1}{2}\right)=\sqrt{3}\)

\(\Leftrightarrow4cos2x.cosx-2cosx=\sqrt{3}\)

\(\Leftrightarrow2cos3x+2cosx-2cosx=\sqrt{3}\)

\(\Leftrightarrow cos3x=\frac{\sqrt{3}}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{18}+\frac{k2\pi}{3}\\x=-\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

Answer 3

3.

ĐKXĐ: ...

\(\frac{1}{cosx}+\frac{1}{2sinx.cosx}=\frac{1}{2sinx.cosx.cos2x}\)

\(\Leftrightarrow2sinx.cos2x+cos2x=1\)

\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)

\(\Leftrightarrow2sinx\left(cos2x-sinx\right)=0\)

\(\Leftrightarrow cos2x-sinx=0\)

\(\Leftrightarrow1-2sin^2x-sinx=0\)

\(\Leftrightarrow2sin^2x+sinx-1=0\)

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

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