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

Toán Lớp 11: Giúp em với Sin6x-√3sin3x=0 √2cos(x-π/6)+sin(2x-π/3)=0

Tháng Hai 13, 2023 Toán học 1 Comment 0 views

Toán Lớp 11: Giúp em với
Sin6x-√3sin3x=0
√2cos(x-π/6)+sin(2x-π/3)=0

Comments ( 1 )

  1. minhhaanh
    minhhaanh
    13 Tháng Hai, 2023 at 3:30 chiều
    Reply
    $a)\quad \sin6x – \sqrt3\sin3x = 0$
    $\Leftrightarrow 2\sin3x\cos3x – \sqrt3\sin3x = 0$
    $\Leftrightarrow \sin3x\left(2\cos3x -\sqrt3\right)= 0$
    $\Leftrightarrow \left[\begin{array}{l}\sin3x = 0\\\cos3x = \dfrac{\sqrt3}{2}\end{array}\right.$
    $\Leftrightarrow \left[\begin{array}{l}3x = k2\pi\\3x = \dfrac{\pi}{6} + k2\pi\\3x= -\dfrac{\pi}{6} + k2\pi\end{array}\right.$
    $\Leftrightarrow \left[\begin{array}{l}x = k\dfrac{2\pi}{3}\\x = \dfrac{\pi}{18} + k\dfrac{2\pi}{3}\\x= -\dfrac{\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)$
    $b)\quad \sqrt2\cos\left(x – \dfrac{\pi}{6}\right) + \sin\left(2x – \dfrac{\pi}{3}\right)= 0$
    $\Leftrightarrow \sqrt2\cos\left(x – \dfrac{\pi}{6}\right) + 2\sin\left(x – \dfrac{\pi}{6}\right)\cos\left(x – \dfrac{\pi}{6}\right) = 0$
    $\Leftrightarrow \cos\left(x – \dfrac{\pi}{6}\right)\left[\sqrt2+ 2\sin\left(x – \dfrac{\pi}{6}\right)\right]= 0$
    $\Leftrightarrow \left[\begin{array}{l}\cos\left(x – \dfrac{\pi}{6}\right)= 0\\\sin\left(x – \dfrac{\pi}{6}\right)= -\dfrac{\sqrt2}{2}\end{array}\right.$
    $\Leftrightarrow \left[\begin{array}{l}x – \dfrac{\pi}{6} =\dfrac{\pi}{2} + k\pi\\x – \dfrac{\pi}{6}= -\dfrac{\pi}{4} + k2\pi\\x – \dfrac{\pi}{6}= \dfrac{5\pi}{4} + k2\pi\end{array}\right.$
    $\Leftrightarrow \left[\begin{array}{l}x=\dfrac{2\pi}{3} + k\pi\\x = -\dfrac{\pi}{12} + k2\pi\\x = \dfrac{17\pi}{12} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

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

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