(cos12x-2cos6x-3) / (12x^2-8pix+pi^2) =0
[tex]sin^2x+sin^2 2x+sin^2 3x +sin^2 4x=2[/tex]
Cảm ơn nhìu.
câu 1:
do mẫu luôn khác 0
pt\Leftrightarrowcos12x - 2cos6x-3=0
\Leftrightarrow2cos^2(6x)-1-2cos6x-3=0
\Leftrightarrowcos^2(6x)-2cos6x-4=0
\Leftrightarrowcos6x=0 hoặc cos6x=1
\Leftrightarrowx=pi/12+kpi/6 hoặc x= kpi/3
Câu 2:
pt\Leftrightarrow(1-cos2x)/2+(1-cos4x)/2+(1-cos6x)/2+(1-cos8x)/2=2
\Leftrightarrowcos2x+cos8x+cos4x+cos6x=0
\Leftrightarrow2cos5xcos2x+2cos5xcosx=0
\Leftrightarrow2cos5x(cos2x+cosx)=0
\Leftrightarrow2cos5x(2cos^2x - 1+cosx)=0
\Leftrightarrow
[tex]\left[\begin{cos5x=0}\\{cosx = 1/2}\\{cosx=-1} [/tex]
\Leftrightarrow
[tex]\left[\begin{x=pi/10+kpi/5}\\{x = pi/3+k2pi or x=-pi/3+k2pi}\\{x=pi+k2pi} [/tex]