a.
[tex]4sinx.sin(\frac{\pi}{3}-x)sin(\frac{\pi}{3}+x)+3cos2x=5\\\Leftrightarrow 4sinx.\frac{-1}{2}(cos\frac{2\pi}{3}-cos2x)+3cos2x=5\\\Leftrightarrow -2sinx(\frac{-1}{2}-cos2x)+3cos2x=5\\\Leftrightarrow 2sinxcos2x+sinx+3cos2x=5\\\Leftrightarrow 4sin^3x+6sin^2x-3sinx+2=0\\\Leftrightarrow 6sin^2x+2-sin3x=0[/tex]
Nhận thấy $VT>0$ nên PTVN
b.
[tex]\Leftrightarrow 1+\frac{8}{3}sin4x=sin^4x+sin^2xcos^2x+cos^4x\\\Leftrightarrow 1+\frac{8}{3}sin4x=sin^2x+cos^4x\\\Leftrightarrow cos^4x-cos^2x-\frac{8}{3}sin4x=0\\\Leftrightarrow cos^2x(1-cos^2x)+\frac{8}{3}sin4x=0\\\ \Leftrightarrow sin^2xcos^2x+\frac{8}{3}sin4x=0\\\Leftrightarrow \frac{1}{4}sin^22x+\frac{16}{3}sin2xcos2x=0\\\Leftrightarrow sin2x(\frac{1}{4}sin2x+\frac{16}{3}cos2x)=0\\\Leftrightarrow ...[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
