cos6x + 3cos8x + 3cos10x + cos12x=0 Tìm x

$\cos6x+\cos12x+3(\cos8x+\cos10x)=0$

$\Leftrightarrow 2\cos9x.\cos3x +6\cos9x.\cos x=0$

$\Leftrightarrow 2\cos9x(\cos3x+3\cos x)=0$

+ TH1: $\cos9x=0\Leftrightarrow x=\dfrac{\pi}{18}+\dfrac{k\pi}{9}$

+ TH2: $\cos3x+3\cos x=0$

$\Leftrightarrow \cos(x+2x)+3\cos x=0$

$\Leftrightarrow \cos x\cos2x-\sin x\sin2x+3\cos x=0$

$\Leftrightarrow \cos x(2\cos^2x-1)-2\sin^2x\cos x+3\cos x=0$

$\Leftrightarrow 2\cos^3x-\cos x-2\cos x(1-\cos^2x)+3\cos x=0$

$\Leftrightarrow 4\cos^3x=0$

$\Leftrightarrow \cos x=0$

$\Leftrightarrow x=\dfrac{\pi}{2}+k\pi$

Vậy $x=\dfrac{\pi}{18}+\dfrac{k\pi}{9}$

Đáp án:

Giải thích các bước giải: cos6x + 3cos8x + 3cos10x + cos12x=0

( cos6x +cos12x)+ 3( cos8x+ cos10x) =0

2cos9x.cos3x +6.cos9x.cosx =0

2cos9x ( cos3x+3cosx) =0

2cos9x ( 4cos^3 x - 3cosx+3cosx) =0

2cos9x . 4 cos^3 x =0

cos9x =0 hoặc cosx =0

x=pi/18 +kpi/9 hoặc x=pi/2 +kpi