cos6x + 3cos8x + 3cos10x + cos12x=0 Tìm x 05/10/2021 Bởi Arya 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 $x=\dfrac{\pi}{18}+\dfrac{k\pi}{9}$ Bình luận
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 Bình luận
$\cos6x+\cos12x+3(\cos8x+\cos10x)=0$
$\Leftrightarrow 2\cos9x.\cos3x +6\cos9x.\cos x=0$
$\Leftrightarrow 2\cos9x(\cos3x+3\cos x)=0$
+ 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