2 replies to this topic

[FrankyLampard]

Giải phương trình:
$1+sin(\dfrac{x}{2}).sinx - cos(\dfrac{x}{2}).sin^2x=2cos^2(\dfrac{\pi}{4} - \dfrac{x}{2})$

[keichan_299]

Giải phương trình:
$1+sin(x/2).sinx - cos(x/2).sin^2x=2cos^2(\pi/4 - x/2)$

$<=> 1+sin \frac{x}{2}sin x- cos \frac{x}{2} sin^{2}x=1+sinx <=> sinx (sin \frac{x}{2}-cos \frac{x}{2}sinx-1)=0$

[thanhelf96]

$sin ( \frac{x}{2} ).sinx-cos ( \frac{x}{2} $
$\Rightarrow sin ( \frac{x}{2} ).sinx-cos ( \frac{x}{2} ).sin^{2}x=cos ( \frac{\Pi }{2}-x )$
$\Rightarrow sin ( \frac{x}{2} ).sinx-cos ( \frac{x}{2} ).sin^{2}x=cos ( \frac{\Pi }{2}-x )=2cos^{2} ( \frac{\Pi }{4} -\frac{x}{2} )-1$
$\Rightarrow sin ( \frac{x}{2} ).sinx-cos ( \frac{x}{2} ).sin^{2}x=sinx$
$\Rightarrow sinx (sin\frac{x}{2} -cos\frac{x}{2}.sinx-1 )=0$
$\Rightarrow sinx = 0$ hoặc $sin\frac{x}{2} -cos\frac{x}{2}.sinx-1=0$
$sin ( \frac{x}{2} )-cos ( \frac{x}{2} ).2sin ( \frac{x}{2} ).cos ( \frac{x}{2} )-1=0$
$\Rightarrow sin ( \frac{x}{2} )-2sin ( \frac{x}{2} ). ( 1-sin^{2} \frac{x}{2} )-1=0$
đến đây bạn tự làm tiếp nha^-^

Edited by hoangtrong2305, 25-07-2012 - 22:23.