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222-9+11+12:2*14+14 = ? ( )

## Toán Lớp 11: sin 3x – $\sqrt{3}$ cos3x=2sinx

Toán Lớp 11: sin 3x – $\sqrt{3}$ cos3x=2sinx

1. Giải đáp:x=π/3+k2π
x=4π/15+k2π/5
Lời giải và giải thích chi tiết:sin3x-3.cos3x=2.sin2x
chia 2 vế cho 2:
1/2.sin3x-√3.cos3x=sin2x
=>cosπ/3.sin3x-sinπ/3.cos3x=sin2x
=>sin(3x-π/3)=sin 2x
<=>3x-π/3=2x+k2π
<=>x=π/3+k2π
x=4π/15+k2π/5

2. Giải đáp:
$S =\left\{\dfrac{\pi}{6} + k\pi;\ \dfrac{\pi}{3} + \dfrac{k\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}$
Lời giải và giải thích chi tiết:
$\quad \sin3x -\sqrt3\cos3x =2\sin x$
$\Leftrightarrow \dfrac12\sin3x – \dfrac{\sqrt3}{2}\cos3x =\sin x$
$\Leftrightarrow \sin\left(3x -\dfrac{\pi}{3}\right)=\sin x$
$\Leftrightarrow \left[\begin{array}{l}3x -\dfrac{\pi}{3}= x + k2\pi\\3x – \dfrac{\pi}{3}= \pi – x + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} + k\pi\\x = \dfrac{\pi}{3} + \dfrac{k2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)$
Vậy $S =\left\{\dfrac{\pi}{6} + k\pi;\ \dfrac{\pi}{3} + \dfrac{k\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}$