please help me ^^^^^^^^^^^^

Question

1)

$\tan \frac{x}{2}.\cos x+\sin 2x=0$

2)

$\sqrt 3\sin \frac{x}{2}-\cos \frac{x}{2}=2\sin 3x$

★★.P.I.N.O.★★ · Answer 1 · 8/15/2015 2:26:42 PM

1. Điều kiện:

$\frac{x}{2}\neq \frac{\pi}{2}+k\pi\Leftrightarrow x\neq\pi +k2\pi;k \in \mathbb Z.$

$\tan \frac{x}{2}.\cos x + \sin 2x=0$

$\Leftrightarrow \tan \frac{x}{2}.\cos x + 2\sin x\cos x=0$

$\Leftrightarrow \left[\ \begin{array}{l} \cos x=0\\ \tan \frac{x}{2}+2\sin x=0(\bigstar) \end{array} \right.\left[\ \begin{array}{l} x=\frac{\pi}{2}+k\pi\\ \tan \frac{x}{2}+2\sin x=0(\bigstar) \end{array} \right.$

$(\bigstar)\Leftrightarrow \sin \frac{x}{2}+2\sin \frac{x}{2}.\cos^{2} \frac{x}{2}=0$

$\Leftrightarrow \left[\ \begin{array}{l} \sin \frac{x}{2}=0\\ 1+2\cos^{2} \frac{x}{2}=0(vn) \end{array} \right.$

$\Leftrightarrow \frac{x}{2}=k\pi\Leftrightarrow x=k.2\pi;k \in \mathbb Z.$

Đối chiếu điều kiện ta được:

$x=\frac{\pi}{2}+k\pi;x=k.2\pi;k \in \mathbb Z.$

2. Phương trình đã cho tương đương với:

$\frac{\sqrt 3}{2}\sin \frac{x}{2}-\frac{1}{2}\cos \frac{x}{2}=\sin 3x$

$\Leftrightarrow \sin (\frac{x}{2}-\frac{\pi}{6})=\sin 3x$

$\Leftrightarrow \left[\ \begin{array}{l} \frac{x}{2}-\frac{\pi}{6}=3x+k.2\pi\\ \frac{x}{2}-\frac{\pi}{6}=\pi-3x+k.2\pi \end{array} \right.\Leftrightarrow \left[\ \begin{array}{l} x=-\frac{\pi}{15}+k.\frac{4\pi}{5}\\ x=\frac{\pi}{3}+k.\frac{4\pi}{7} \end{array} \right.(k \in \mathbb Z)$

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