Giải phương trình \(\cos x \cos \frac{x}{2} \cos \frac{3 x}{2}-\sin x \sin \frac{x}{2} \sin \frac{3 x}{2}=\frac{1}{2}\)
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Lời giải:
Báo sai\(\begin{aligned} &\cos x \cos \frac{x}{2} \cos \frac{3 x}{2}-\sin x \sin \frac{x}{2} \sin \frac{3 x}{2}=\frac{1}{2} \\ &\Leftrightarrow \frac{1}{2} \cos x(\cos 2 x+\cos x)+\frac{1}{2} \sin x(\cos 2 x-\cos x)=\frac{1}{2} \\ &\Leftrightarrow \cos x \cos 2 x+\cos ^{2} x+\sin x \cos 2 x-\sin x \cos x=1 \\ &\Leftrightarrow \cos 2 x(\sin x+\cos x)+1-\sin ^{2} x-\sin x \cos x-1=0 \\ &\Leftrightarrow \cos 2 x(\sin x+\cos x)-\sin x(\sin x+\cos x)=0 \\ &\Leftrightarrow(\sin x+\cos x)(\cos 2 x-\sin x)=0 \\ &\Leftrightarrow(\sin x+\cos x)\left(-2 \sin ^{2} x-\sin x+1\right)=0 \end{aligned}\)
\(\Leftrightarrow\left[\begin{array}{c} \sin x+\cos x=0 \\ 2 \sin ^{2} x+\sin x-1=0 \end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{c} \tan x=-1 \\ \sin x=-1 \\ \sin x=1 / 2 \end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{c} x=-\frac{\pi}{4}+k \pi \\ x=-\frac{\pi}{2}+k 2 \pi \\ x=\frac{\pi}{6}+k 2 \pi \vee x=\frac{5 \pi}{6}+k 2 \pi \end{array}\right.\)
