Question

Câu 2: Giải phương trình \(2 \sin \left( 2x - \frac{\pi}{4} \right) + \sqrt{3} = 0\)

Answer 1

Ta có: \(2\cdot\sin\left(2x-\frac{\pi}{4}\right)+\sqrt3=0\)

=>\(2\cdot\sin\left(2x-\frac{\pi}{4}\right)=-\sqrt3\)

=>\(\sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt3}{2}\)

=>\(\left[\begin{array}{l}2x-\frac{\pi}{4}=-\frac{\pi}{3}+k2\pi\\ 2x-\frac{\pi}{4}=\pi+\frac{\pi}{3}+k2\pi=\frac43\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}2x=-\frac{\pi}{3}+\frac{\pi}{4}+k2\pi=-\frac{1}{12}\pi+k2\pi\\ 2x=\frac43\pi+\frac{\pi}{4}+k2\pi=\frac{19}{12}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac{1}{24}\pi+k\pi\\ x=\frac{19}{24}\pi+k\pi\end{array}\right.\)