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

Toán Lớp 11: nghiệm của pt sin^2 x+căn3 sinx cosx=1

Toán Lớp 11: nghiệm của pt sin^2 x+căn3 sinx cosx=1

Comments ( 1 )

  1. Giải đáp:
    \(S = \left\{\dfrac{\pi}{6} +k\pi;\ \dfrac{\pi}{2} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}\) 
    Lời giải và giải thích chi tiết:
    \(\begin{array}{l}
    \quad \sin^2x + \sqrt3\sin x\cos x = 1\\
    \Leftrightarrow \dfrac{1 – \cos2x}{2} + \dfrac{\sqrt3}{2}\sin2x = 1\\
    \Leftrightarrow \dfrac{\sqrt3}{2}\sin2x – \dfrac12\cos2x = \dfrac12\\
    \Leftrightarrow \sin\left(2x – \dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{6}\\
    \Leftrightarrow \left[\begin{array}{l}2x – \dfrac{\pi}{6} = \dfrac{\pi}{6}  +k2\pi\\2x – \dfrac{\pi}{6} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\
    \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} +k\pi\\x = \dfrac{\pi}{2} + k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
    \text{Vậy}\ S = \left\{\dfrac{\pi}{6} +k\pi;\ \dfrac{\pi}{2} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}
    \end{array}\)

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