Toán Lớp 11: Giải pt
a)sinx-căn3 cosx=2sin3x
b)cos2x+căn3 sin2x-2cosx=0
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Toán Lớp 11: Giải pt
a)sinx-căn3 cosx=2sin3x
b)cos2x+căn3 sin2x-2cosx=0
Comments ( 2 )
a)\quad S =\left\{-\dfrac{\pi}{6} + k\pi;\ \dfrac{\pi}{3} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\\
b)\quad S =\left\{- \dfrac{\pi}{3} +k2\pi;\ – \dfrac{\pi}{9} + k\dfrac{2\pi}{3}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)
a)\quad \sin x – \sqrt3\cos x =2\sin3x\\
\Leftrightarrow \dfrac12\sin x – \dfrac{\sqrt3}{2}\cos x = \sin3x\\
\Leftrightarrow \sin\left(x – \dfrac{\pi}{3}\right) = \sin3x\\
\Leftrightarrow \left[\begin{array}{l}x – \dfrac{\pi}{3} = 3x + k2\pi\\x – \dfrac{\pi}{3} = \pi – 3x + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x =-\dfrac{\pi}{6} + k\pi\\x = \dfrac{\pi}{3} + k\dfrac{\pi}{2}\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S =\left\{-\dfrac{\pi}{6} + k\pi;\ \dfrac{\pi}{3} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\\
b)\quad \cos2x – \sqrt3\sin2x – 2\cos x = 0\\
\Leftrightarrow \dfrac12\cos2x- \dfrac{\sqrt3}{2}\sin2x = \cos x\\
\Leftrightarrow \cos\left(2x + \dfrac{\pi}{3}\right) =\cos x \\
\Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{3} = x + k2\pi\\2x + \dfrac{\pi}{3} = -x + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{3} +k2\pi\\x = – \dfrac{\pi}{9} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S =\left\{- \dfrac{\pi}{3} +k2\pi;\ – \dfrac{\pi}{9} + k\dfrac{2\pi}{3}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)