$a)\,2\left(\cos x+\sqrt{3}\sin x\right)\cos x=\cos x-\sqrt{3}\sin x+1\\b)\,\dfrac{\sin x\sin2x+2\sin x\cos^2x+\sin x\cos x}{\cos\left(x-\dfrac{\pi}{4}\right)}=\sqrt{6}\cos2x\\c)\,\sin\left(\dfrac{3\pi}{10}-\dfrac{x}{2}\right)=\dfrac{1}{2}\sin\left(\dfrac{\pi}{10}+\dfrac{3x}{2}\right)$ 
c)
Phương trình tương đương với:
$\sin\left(\frac{3\pi}{10}-\frac{x}{2}\right)=\frac{1}{2}\sin\left (\frac{9\pi}{10}-\frac{3x}{2} \right )$
Đặt: $\frac{3\pi}{10}-\frac{x}{2}=t$, ta có:
      $\sin t=\frac{1}{2}\sin3t$
$\Leftrightarrow 2\sin t=3\sin t-4\sin^3t$
$\Leftrightarrow \left[\begin{array}{l} \sin t=0\\\sin t=\frac{1}{2}\\\sin t=\frac{-1}{2} \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} t=k\pi\\t=\pm\frac{\pi}{6}+k2\pi\\t=\pm\frac{5\pi}{6}+k2\pi \end{array} \right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[ \begin{array}{l} x=\frac{3\pi}{5}-k\pi\\x=\frac{4\pi}{15}-k4\pi\\x=\frac{14\pi}{15}-k4\pi \end{array} \right.,k\in\mathbb{Z}$
Anh Khang ơi! :) –  Xusint 18-10-12 08:06 PM
a)
Phương trình tương đương với:
     $2\cos^2x-1+2\sqrt3\sin x\cos x=\cos x-\sqrt3\sin x$
$\Leftrightarrow \frac{1}{2}\cos2x+\frac{\sqrt3}{2}\sin2x=\frac{1}{2}\cos x-\frac{\sqrt3}{2}\sin x$
$\Leftrightarrow \cos\left(2x-\frac{\pi}{3}\right)=\cos\left(x+\frac{\pi}{3}\right)$
$\Leftrightarrow \left[ \begin{array}{l} 2x-\frac{\pi}{3}=x+\frac{\pi}{3}+k2\pi\\2x-\frac{\pi}{3}=-x-\frac{\pi}{3}+k2\pi \end{array} \right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[ \begin{array}{l} x=\frac{2\pi}{3}+k2\pi\\x=\frac{k2\pi}{3} \end{array} \right.,k\in\mathbb{Z} \Leftrightarrow x=\frac{k2\pi}{3},k\in\mathbb{Z}$
chuyển 1 sang bên VT bạn nhá, mình sửa lại rồi đó. –  khangnguyenthanh 22-10-12 09:02 PM
Anh Khang ơi bài này anh giải thiếu rồi ạ bên VP đề bài còn số 1 nữa mà anh :D –  Xusint 22-10-12 08:55 PM
Anh Khang xem giúp em câu b) với ạ. –  Xusint 17-10-12 12:00 PM

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