3) [tex]\frac{1}{2}\left ( cos6x+cos4x \right )+\frac{1}{2}(cos8x-cos4x)=cos7x<=>cos6x+cos8x=2cos7x<=>cos8x-cos7x=cos7x-cos6x<=>-2sin\frac{x}{2}sin\frac{15x}{2}=-2sin\frac{x}{2}sin\frac{13x}{2}<=>sin\frac{x}{2}=0 \vee sin\frac{15x}{2}=sin\frac{13x}{2}<=>2cos7xsin\frac{x}{2}=0=>cos7x=0[/tex]
2) [tex]cos^3x(4cos^3x-3cosx)+sin^3x(3sinx-4sin^3x)=\frac{\sqrt{2}}{4}<=>4cos^6x-3cos^4x+3sin^4x-4sin^6x=\frac{\sqrt{2}}{4}<=>4(cos^2x-sin^2x)(cos^4x+cos^2xsin^2x+sin^4x)+3(sin^2x-cos^2x)=\frac{\sqrt{2}}{4}<=>4cos^2x-sin^2x)(1-cos^2xsin^2x)+3(sin^2x-cos^2x)=\frac{\sqrt{2}}{4}<=>4cos2x(1-\frac{1}{4}sin^22x)-3cos2x=\frac{\sqrt{2}}{4}<=>4cos2x(\frac{3}{4}+\frac{1}{4}cos^22x)-3cos2x=\frac{\sqrt{2}}{4}<=>cos^32x=\frac{\sqrt{2}}{4}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
