Câu hỏi của Nguyễn Bích Khuê

Question

(cosx+sin2x)/(cos2x+sinx)=căn3

Hung nguyen · Accepted Answer

$\dfrac{cosx+sin2x}{cos2x+sinx}=\sqrt{3}$

$\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x-\dfrac{1}{2}sin2x$

$\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos\left(2x+\dfrac{\pi}{6}\right)$

Làm nốt nhéCâu trả lời: $\dfrac{cosx+sin2x}{cos2x+sinx}=\sqrt{3}$

$\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x-\dfrac{1}{2}sin2x$

$\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos\left(2x+\dfrac{\pi}{6}\right)$

Làm nốt nhé