Giúp em giải mấy pt này với, cảm ơn ạ!

Question

a,

$\frac{\cos ^{2}x\left ( \cos x - 1\right ) }{\sin x + \cos x}$ =

$2 \left ( 1 + \sin x \right )$

b,

$\sin x + \sin 2x + \sin 3x - \sqrt{3}\left ( \cos x +\cos 2x + \cos 3x \right )$ = 0

score 1 · Answer 1

b)
pt

$\Leftrightarrow sin2x+(sinx+sin3x)-\sqrt{3}(cos2x+cosx+cos3x)=0$

$\Leftrightarrow sin2x+2sin2x.cosx-\sqrt{3}(cos2x+2cos2x.cosx)=0$

$\Leftrightarrow sin2x(2cosx+1)-\sqrt{3}cos2x(2cosx+1)=0$

$\Leftrightarrow (2cosx+1)(sin2x-\sqrt{3}cos2x)=0$

$\Leftrightarrow \left[ {\begin{matrix} cosx=-1/2(*)\\ sin2x-\sqrt{3}cos2x=0 (**) \end{matrix}} \right.$
+)

$(*)\Leftrightarrow x=\pm \frac{2\pi}{3}+k2\pi (k\in Z)$
+)

$(**)\Leftrightarrow \frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x=0\Leftrightarrow sin(2x-\frac{\pi}{3})=0\Leftrightarrow 2x-\frac{\pi}{3}=k\pi$

$\Leftrightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}$ (

$k\in Z$ )

Vậy pt có 3 họ nghiệm

score 2 · Answer 2

a. ĐK:

$\sin x+\cos x\ne0 \Leftrightarrow x\ne\dfrac{-\pi}{4}+k\pi,k\in\mathbb{Z}$

Phương trình đã cho tương đương với:

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

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

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

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

$\Leftrightarrow \left[\begin{array}{l}\sin x=-1\\\cos x=-1\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x=-\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{array}\right.,k\in\mathbb{Z}$

em cảm ơn nhiều ạ, – . 19-08-14 09:41 AM

2 Đáp án