Question

Tổng các nghiệm của phương trình \(cos\left(2x+\dfrac{\pi}{6}\right)+sin\left(\dfrac{\pi}{3}-2x\right)=\sqrt{2}\) thuộc (0;π) là?

 

Answer 1

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

=>cos(2x+pi/6)=căn 2/2

=>2x+pi/6=pi/4+k2pi hoặc 2x+pi/6=-pi/4+k2pi

=>2x=1/2pi+k2pi hoặc 2x=-5/12pi+2kpi

=>x=1/4pi+kpi hoặc x=-5/24pi+kpi

Answer 2

Đặt \(2x+\dfrac{\pi}{6}=t\Rightarrow\dfrac{\pi}{3}-2x=\dfrac{\pi}{2}-t\)

\(\Rightarrow cost+sin\left(\dfrac{\pi}{2}-t\right)=\sqrt{2}\)

\(\Leftrightarrow2cost=\sqrt{2}\)

\(\Rightarrow cost=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow cos\left(2x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{4}+k2\pi\\2x+\dfrac{\pi}{6}=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{24}+k\pi\\x=-\dfrac{5\pi}{24}+k\pi\end{matrix}\right.\)

\(\Rightarrow x=\left\{\dfrac{\pi}{24};\dfrac{19\pi}{24}\right\}\)