Phương trình: \(2 \sqrt{3} \sin \left(x-\frac{\pi}{8}\right) \cos \left(x-\frac{\pi}{8}\right)+2 \cos ^{2}\left(x-\frac{\pi}{8}\right)=\sqrt{3}+1\) có nghiệm là:
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Lời giải:
Báo sai\(\begin{array}{l} 2 \sqrt{3} \sin \left(x-\frac{\pi}{8}\right) \cos \left(x-\frac{\pi}{8}\right)+2 \cos ^{2}\left(x-\frac{\pi}{8}\right)=\sqrt{3}+1 \Leftrightarrow \sqrt{3} \sin \left(2 x-\frac{\pi}{4}\right)+\cos \left(2 x+\frac{\pi}{4}\right)+1=\sqrt{3}+1 \\ \Leftrightarrow \frac{\sqrt{3}}{2} \sin \left(2 x-\frac{\pi}{4}\right)+\frac{1}{2} \cos \left(2 x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2} \Leftrightarrow \sin \frac{\pi}{3} \cdot \sin \left(2 x-\frac{\pi}{4}\right)+\cos \frac{\pi}{3} \cdot \cos \left(2 x-\frac{\pi}{4}\right)=\cos \frac{\pi}{6} \\ \Leftrightarrow \cos \left(2 x-\frac{\pi}{4}-\frac{\pi}{3}\right)=\cos \frac{\pi}{6} \cdot \Leftrightarrow\left[\begin{array}{c} 2 x-\frac{7 \pi}{12}=\frac{\pi}{6}+k 2 \pi \\ 2 x-\frac{7 \pi}{12}=-\frac{\pi}{6}+k 2 \pi \end{array} \Leftrightarrow\left[\begin{array}{c} x=\frac{3 \pi}{8}+k \pi \\ 5=\frac{5 \pi}{12}+k \pi \end{array}, k \in \mathbb{Z}\right.\right. \end{array}\)
