Câu hỏi của Hoang Hon

Question

Nguyễn Minh Quang · Accepted Answer

ta có :

$\hept{\begin{cases}5sin^2x+3cos^2x=3+2sin^2x=4-cos2x\2\sqrt{3}sinx.cosx=\sqrt{3}sin2x\end{cases}}$

Nên ta có : $4-cos2x+\sqrt{3}sin2x=2\Leftrightarrow-cos2x+\sqrt{3}sin2x=-2$

$\Leftrightarrow-\frac{1}{2}cós2x+\frac{\sqrt{3}}{2}sin2x=-1\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=-1$

$\Leftrightarrow2x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\Leftrightarrow x=-\frac{\pi}{6}+k\pi$

Câu trả lời:

ta có :

$\hept{\begin{cases}5sin^2x+3cos^2x=3+2sin^2x=4-cos2x\2\sqrt{3}sinx.cosx=\sqrt{3}sin2x\end{cases}}$

Nên ta có : $4-cos2x+\sqrt{3}sin2x=2\Leftrightarrow-cos2x+\sqrt{3}sin2x=-2$

$\Leftrightarrow-\frac{1}{2}cós2x+\frac{\sqrt{3}}{2}sin2x=-1\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=-1$

$\Leftrightarrow2x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\Leftrightarrow x=-\frac{\pi}{6}+k\pi$