Inequality . HELP me

**mp3qz** · July 21st 2009, 07:27 AM

If you can, please with this trigonometric equation

Worried)
2./

$\frac{2sin2x+2cosx-2sinx-1}{2cosx-1}=cos2x+\sqrt{3}(sinx+1)$

THANKS YOU VERY MUCH

**red_dog** · July 21st 2009, 07:59 AM

2) $2\cos x-1\neq 0\Rightarrow x\neq\pm\frac{\pi}{3}+2k\pi, \ k\in\mathbb{Z}$

The left side member is:

$\frac{4\sin x\cos x+2\cos x-2\sin x-1}{2\cos x-1}=\frac{(2\cos x-1)(2\sin x+1)}{2\cos x-1}=2\sin x+1$

Then the equation becomes

$2\sin x+1=\cos 2x+\sqrt{3}(\sin x+1)\Rightarrow 2\sin^2x+(2-\sqrt{3})\sin x-\sqrt{3}=0$

$\sin x=-1\Rightarrow x=-\frac{\pi}{2}+2k\pi, \ k\in\mathbb{Z}$

$\sin x=\frac{\sqrt{3}}{2}\Rightarrow x=-\frac{\pi}{3}+(2k+1)\pi, \ k\in\mathbb{Z}$

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