Question

\(\cos2x=\sin\left(x+\frac{\Pi}{5}\right)\)

Answer 1

Lời giải:

$\cos 2x=\sin (x+\frac{\pi}{5})=\cos [\frac{\pi}{2}-(x+\frac{\pi}{5})]=\cos (\frac{3\pi}{10}-x)$

$\Leftrightarrow \cos 2x-\cos (\frac{3\pi}{10}-x)=0$

$\Leftrightarrow -2\sin (\frac{x}{2}+\frac{3\pi}{20})\sin (\frac{3}{2}x-\frac{3\pi}{20})=0$

$\Rightarrow \sin (\frac{x}{2}+\frac{3\pi}{20})=0$ hoặc $\sin \frac{3}{2}\pi-\frac{3\pi}{20})=0$

$\Rightarrow x=2k\pi-\frac{3}{10}\pi$ với $k$ nguyên hoặc $x=\frac{\pi}{10}+\frac{2}{3}k\pi$ với $k$ nguyên