Anh ơi có phải cách giải như thế này không ạ ^ ^$cos(\frac{x}{2} + \frac{\pi}{3}) sin(\frac{3x}{2}-\frac{\pi}6) = -\frac{1}{4}$$ \Leftrightarrow \frac{\sqrt{3}}{4}(cos\frac{x}{2}.\frac{1}{2}-sin\frac{x}{2}. \sqrt{3}/2)(sin\frac{3x}{2}.\sqrt{3}/2-cos\frac{3x}{2}.\frac{1}{2})= -\frac{1}{4}$$\Leftrightarrow \sqrt{3}/4(cos\frac{x}{2}.sin\frac{3x}{2}+sin\frac{x}{2}.cos\frac{3x}{2})-\frac{3}{4}sin\frac{x}{2}sin\frac{3x}{2}-\frac{1}{4}cos\frac{x}{2}cos\frac{3x}{2}=-1/4$$\Leftrightarrow \sqrt{3}/4.sin2x-1/8(cos2x+cosx)+3/8(cos2x-cosx)=-1/4$$\Leftrightarrow \sqrt{3}/4 sin2x +1/4cos2x -1/2cosx=-1/4$$\Leftrightarrow \sqrt{3}/2.2sinxcosx +1/2cos2x+1/2-cosx=0$$\Leftrightarrow \sqrt{3}sinxcosx + cos^2x-cosx=0$$\Leftrightarrow 3sin^2xcos^2x= (cos^2x-cosx)^2=0$$\Leftrightarrow 4cos^4x -2cos^3x-2cos^2x=0$$\Leftrightarrow 2cos^2(2cos^2x-cosx-1)=0$$\Leftrightarrow cosx=0 khi x=\frac{\pi}{2} hay cos=1 khi x=0 và cos=-1/2 khi x=\frac{2\pi}{3}$
Anh ơi có phải cách giải như thế này không ạ ^ ^$cos(\frac{x}{2} + \frac{\pi}{3}) sin(\frac{3x}{2}-\frac{\pi}6) = -\frac{1}{4}$$ \Leftrightarrow \frac{\sqrt{3}}{4}(cos\frac{x}{2}.\frac{1}{2}-sin\frac{x}{2}. \sqrt{3}/2)(sin\frac{3x}{2}.\sqrt{3}/2-cos\frac{3x}{2}.\frac{1}{2})= -\frac{1}{4}$$\Leftrightarrow \sqrt{3}/4(cos\frac{x}{2}.sin\frac{3x}{2}+sin\frac{x}{2}.cos\frac{3x}{2})-\frac{3}{4}sin\frac{x}{2}sin\frac{3x}{2}-\frac{1}{4}cos\frac{x}{2}cos\frac{3x}{2}=-\frac{1}{4}$$\Leftrightarrow \sqrt{3}/4.sin2x-1/8(cos2x+cosx)+3/8(cos2x-cosx)=-1/4$$\Leftrightarrow \sqrt{3}/4 sin2x +1/4cos2x -1/2cosx=-1/4$$\Leftrightarrow \sqrt{3}/2.2sinxcosx +1/2cos2x+1/2-cosx=0$$\Leftrightarrow \sqrt{3}sinxcosx + cos^2x-cosx=0$$\Leftrightarrow 3sin^2xcos^2x= (cos^2x-cosx)^2=0$$\Leftrightarrow 4cos^4x -2cos^3x-2cos^2x=0$$\Leftrightarrow 2cos^2(2cos^2x-cosx-1)=0$$\Leftrightarrow cosx=0 khi x=\frac{\pi}{2} hay cos=1 khi x=0 và cos=-1/2 khi x=\frac{2\pi}{3}$