Toán Lớp 11: Giải pt
a)sin3x -căn3 cos3x=2sin2x
b)sin9x-cos5x=căn3(sin5x+cos9x)
c)sin7x-cos2x=căn3(sin2x-cos7x)
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Toán Lớp 11: Giải pt
a)sin3x -căn3 cos3x=2sin2x
b)sin9x-cos5x=căn3(sin5x+cos9x)
c)sin7x-cos2x=căn3(sin2x-cos7x)
Comments ( 2 )
a)\quad S = \left\{\dfrac{\pi}{3} + k2\pi;\ \dfrac{4\pi}{15} + k\dfrac{2\pi}{5}\ \Bigg|\ k\in\Bbb Z\right\}\\
b)\quad S = \left\{\dfrac{\pi}{8} + k\dfrac{\pi}{2};\ \dfrac{\pi}{12} + k\dfrac{\pi}{7}\ \Bigg|\ k\in\Bbb Z\right\}\\
c)\quad S = \left\{- \dfrac{\pi}{30} + k\dfrac{2\pi}{5};\ \dfrac{\pi}{18} + k\dfrac{2\pi}{9}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)
a)\quad \sin3x – \sqrt3\cos3x= 2\sin2x\\
\Leftrightarrow \dfrac{1}{2}\sin3x – \dfrac{\sqrt3}{2}\cos3x= \sin2x\\
\Leftrightarrow \sin\left(3x – \dfrac{\pi}{3}\right) = \sin2x\\
\Leftrightarrow \left[\begin{array}{l}3x – \dfrac{\pi}{3} = 2x + k2\pi\\3x – \dfrac{\pi}{3} = \pi – 2x + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{3} + k2\pi\\x = \dfrac{4\pi}{15} + k\dfrac{2\pi}{5}\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\dfrac{\pi}{3} + k2\pi;\ \dfrac{4\pi}{15} + k\dfrac{2\pi}{5}\ \Bigg|\ k\in\Bbb Z\right\}\\
b)\quad \sin9x – \cos5x = \sqrt3(\sin5x + \cos9x)\\
\Leftrightarrow \sin9x – \sqrt3\cos9x = \sqrt3\sin5x + \cos5x\\
\Leftrightarrow \dfrac{1}{2}\sin9x – \dfrac{\sqrt3}{2}\cos9x = \dfrac{\sqrt3}{2}\sin5x + \dfrac12\cos5x\\
\Leftrightarrow \sin\left(9x – \dfrac{\pi}{3}\right) = \sin\left(5x + \dfrac{\pi}{6}\right)\\
\Leftrightarrow \left[\begin{array}{l}9x – \dfrac{\pi}{3} = 5x + \dfrac{\pi}{6} + k2\pi\\9x – \dfrac{\pi}{3} = \dfrac{5\pi}{6} – 5x +k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{8} + k\dfrac{\pi}{2}\\x = \dfrac{\pi}{12} + k\dfrac{\pi}{7}\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\dfrac{\pi}{8} + k\dfrac{\pi}{2};\ \dfrac{\pi}{12} + k\dfrac{\pi}{7}\ \Bigg|\ k\in\Bbb Z\right\}\\
c)\quad \sin7x – \cos2x = \sqrt3(\sin2x – \cos7x)\\
\Leftrightarrow \sin7x + \sqrt3\cos7x = \sqrt3\sin2x + \cos2x\\
\Leftrightarrow \dfrac{1}{2}\sin7x + \dfrac{\sqrt3}{2}\cos7x = \dfrac{\sqrt3}{2}\sin2x + \dfrac{1}{2}\cos2x\\
\Leftrightarrow \sin\left(7x + \dfrac{\pi}{3}\right) = \sin\left(2x + \dfrac{\pi}{6}\right)\\
\Leftrightarrow \left[\begin{array}{l}7x + \dfrac{\pi}{3} = 2x + \dfrac{\pi}{6} + k2\pi\\7x + \dfrac{\pi}{3} = \dfrac{5\pi}{6} – 2x + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{30} + k\dfrac{2\pi}{5}\\x = \dfrac{\pi}{18} + k\dfrac{2\pi}{9}\end{array}\right.\quad (k\in\Bbb )\\
\text{Vậy}\ S = \left\{- \dfrac{\pi}{30} + k\dfrac{2\pi}{5};\ \dfrac{\pi}{18} + k\dfrac{2\pi}{9}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)