Question

Giải phương trình lượng giác sau

\(2\sqrt{3}sin\left(x-\dfrac{\pi}{8}\right)cos\left(x-\dfrac{\pi}{8}\right)+2sin^2\left(x-\dfrac{\pi}{8}\right)=\sqrt{3}+1\)

 

Answer 1

\(\Leftrightarrow\sqrt{3}sin\left(2x-\dfrac{\pi}{4}\right)+1-cos\left(2x-\dfrac{\pi}{4}\right)=\sqrt{3}+1\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin\left(2x-\dfrac{\pi}{4}\right)-\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{5\pi}{12}=\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{5\pi}{12}=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3\pi}{8}+k\pi\\x=\dfrac{13\pi}{24}+k\pi\end{matrix}\right.\)