Giải phương trình cosxcosx/2cos3x/2-sinxsinx/2sin3x/2=1/2. A. x=-pi/4+kpi, x=pi/6+k2pi, x=5pi/6+k2pi, x=-pi/2+k2pi (k thuôc Z)

* Hướng dẫn giải

$\cos x\cos \frac{x}{2}\cos \frac{3x}{2}-\sin x\sin \frac{x}{2}\sin \frac{3x}{2}=\frac{1}{2}$

$\iff \frac{1}{2}\cos x\left(\cos 2x+\cos x\right)+\frac{1}{2}\sin x\left(\cos 2x-\cos x\right)=\frac{1}{2}$

$\iff \cos x\cos 2x+{\cos }^{2}x+\sin x\cos 2x-\sin x\cos x=1$

$\iff \cos 2x\left(\sin x+\cos x\right)-\sin x\cos x+{\cos }^{2}x-1=0$

$\iff \cos 2x\left(\sin x+\cos x\right)-\sin x\cos x-si{n}^{2}x=0$

$\iff \cos 2x\left(\sin x+\cos x\right)-\sin x\left(\sin x+\cos x\right)=0$

$\iff \left(\sin x+\cos x\right)\left(\cos 2x-\sin x\right)=0$

$\iff \left[\begin{array}{c}\sin x+\cos x=0\\ \cos 2x-\sin x=0\end{array}\right.$

$\iff \left[\begin{array}{c}\sin x=-\cos x\\ 1-2{\sin }^{2}x-\sin x=0\end{array}\right.$

$\iff \left[\begin{array}{c}\tan x=-1\\ \sin x=\frac{1}{2}\\ \sin x=-1\end{array}\right.$

$\iff \left[\begin{array}{c}x=-\frac{\pi }{4}+k\pi \\ x=\frac{\pi }{6}+k2\pi \\ x=\frac{5\pi }{6}+k2\pi \\ x=-\frac{\pi }{2}+k2\pi \end{array}\right.\left(k\in Z\right)$