Giải phương trình $\cos 2 x+7 \cos x-\sqrt{3}(\sin 2 x-7 \sin x)=8$.

Q: Giải phương trình $\cos 2 x+7 \cos x-\sqrt{3}(\sin 2 x-7 \sin x)=8$.

$\begin{array}{l}\text {} \cos 2 x+7 \cos x-\sqrt{3}(\sin 2 x-7 \sin x)=8 \text {. ~(1)} \\\text { (1) } \Leftrightarrow(\cos 2 x-\sqrt{3} \sin 2 x)+7(\cos x+\sqrt{3} \sin x)=8 \\\Leftrightarrow \cos \left(2 x+\frac{\pi}{3}\right)+7 \sin \left(x+\frac{\pi}{6}\right)-4=0 \\\Leftrightarrow 1-2 \sin ^{2}\left(x+\frac{\pi}{6}\right)+7 \sin \left(x+\frac{\pi}{6}\right)-4=0 \\\Leftrightarrow 2 \sin ^{2}\left(x+\frac{\pi}{6}\right)-7 \sin \left(x+\frac{\pi}{6}\right)+3=0 \\\Leftrightarrow\left[\begin{array}{l}\sin \left(x+\frac{\pi}{6}\right)=\frac{1}{2} \\\sin \left(x+\frac{\pi}{6}\right)=3(p t v n)\end{array}\right. \\\Leftrightarrow\left[\begin{array}{l}x=k 2 \pi \\x=\frac{2 \pi}{3}+k 2 \pi\end{array}(k \in \mathbb{Z}) .\right. \\\end{array}$Vậy phương trình có nghiệm $x=k 2 \pi, x=\frac{2 \pi}{3}+k 2 \pi, k \in \mathbb{Z}$.

Question

Accepted Answer

$\begin{array}{l}\text {} \cos 2 x+7 \cos x-\sqrt{3}(\sin 2 x-7 \sin x)=8 \text {. ~(1)} \\text { (1) } \Leftrightarrow(\cos 2 x-\sqrt{3} \sin 2 x)+7(\cos x+\sqrt{3} \sin x)=8 \\Leftrightarrow \cos \left(2 x+\frac{\pi}{3}\right)+7 \sin \left(x+\frac{\pi}{6}\right)-4=0 \\Leftrightarrow 1-2 \sin ^{2}\left(x+\frac{\pi}{6}\right)+7 \sin \left(x+\frac{\pi}{6}\right)-4=0 \\Leftrightarrow 2 \sin ^{2}\left(x+\frac{\pi}{6}\right)-7 \sin \left(x+\frac{\pi}{6}\right)+3=0 \\Leftrightarrow\left[\begin{array}{l}\sin \left(x+\frac{\pi}{6}\right)=\frac{1}{2} \\sin \left(x+\frac{\pi}{6}\right)=3(p t v n)\end{array}\right. \\Leftrightarrow\left[\begin{array}{l}x=k 2 \pi \x=\frac{2 \pi}{3}+k 2 \pi\end{array}(k \in \mathbb{Z}) .\right. \\end{array}$

Vậy phương trình có nghiệm $x=k 2 \pi, x=\frac{2 \pi}{3}+k 2 \pi, k \in \mathbb{Z}$.

Đề thi HSG (CT) 18-19 - Nghệ An - MĐ 6577