Toán Lớp 11: Giải pt:căn3sin2x + cos2x – 2cosx + 1=0
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Toán Lớp 11: Giải pt:căn3sin2x + cos2x – 2cosx + 1=0
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\quad \sqrt3\sin2x + \cos2x + 2\cos x + 1 =0\\
\Leftrightarrow 2\sqrt3\sin x\cos x + (2\cos^2x – 1) + 2\cos x + 1 =0\\
\Leftrightarrow \cos x\left(\sqrt3\sin x + \cos x + 1\right)= 0\\
\Leftrightarrow 2\cos x\left(\dfrac{\sqrt3}{2}\sin x + \dfrac12\cos x + \dfrac12\right)= 0\\
\Leftrightarrow \cos x\left[\sin\left(x + \dfrac{\pi}{6}\right) + \dfrac12\right] =0\\
\Leftrightarrow \left[\begin{array}{l}\cos x =0\\\sin\left(x + \dfrac{\pi}{6}\right)
= -\dfrac12\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\x + \dfrac{\pi}{6} = – \dfrac{\pi}{6} + k2\pi\\x + \dfrac{\pi}{6} = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\x = – \dfrac{\pi}{3} + k2\pi\\x = \pi + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\dfrac{\pi}{2} + k\pi;\ – \dfrac{\pi}{3} + k2\pi;\ \pi + k2\pi\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)