$ocs x + cos 3x+ 2cos 5x = 0$
⇔ $(cos5x+ cos x ) + (cos 5 x + cos 3x ) = 0
⇔ $2cos3x . cos2x + 2 cos 4x . cosx = 0 $
⇔ $(4cos^{3}x-3cosx) cos2x + cos4x.cosx = 0$
⇔ $cosx[(4cos^{2}x-3cosx)cos2x+ cos4x] = 0$
⇔ $cosx[(2cos2x-1)cos2x+2cos^{2}2x-1]=0$
⇔ $cosx(2cos^{2}2x-cos2x-1)=0$
⇔ \(\left[ \begin{array}{l}cosx=0\\cos2x=\frac{1±\sqrt[]{17}}{8} \end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{π}{2}+kπ\\x=±\frac{1}{2}arccos\frac{1±\sqrt[]{17} }{8}+k2π \end{array} \right.\)
Vậy $m=\frac{1±\sqrt[]{17}}{8}$
Chọn $A$
Nhớ cho mik ctlhn nhé!!
