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

Toán Lớp 11: Giải phương trìn: a, cos2x – sinx = 0 b, sin2x + sinx =0 c, sin2x = 0 d, cos2x = 0 e, sin3x = -1 f, cos 3x = -1 g, sin 2x =1/5 h, cosx

Tháng Một 21, 2023 Toán học 1 Comment 0 views

Toán Lớp 11: Giải phương trìn:
a, cos2x – sinx = 0
b, sin2x + sinx =0
c, sin2x = 0
d, cos2x = 0
e, sin3x = -1
f, cos 3x = -1
g, sin 2x =1/5
h, cosx = -1/2

Comments ( 1 )

  1. kimnguenoanh
    kimnguenoanh
    21 Tháng Một, 2023 at 6:11 chiều
    Reply
    Giải đáp:
    $a) \left[\begin{array}{cc}  x  =  \dfrac{\pi}{6} + k 2 \pi & ( k \in \mathbb{Z}) \\x  =  \dfrac{5\pi}{6} + l 2 \pi & ( l \in \mathbb{Z})  \\ x  =  -\dfrac{\pi}{2} + m 2 \pi & (m \in \mathbb{Z})  \end{array} \right.\\ b)\left[\begin{array}{cc} x = k \pi & ( k \in \mathbb{Z})  \\ x=\dfrac{2\pi}{3} + l 2 \pi & ( l \in \mathbb{Z}) \\ x=\dfrac{-2\pi}{3} + m 2 \pi & ( m\in \mathbb{Z})\end{array} \right.\\ c)x=\dfrac{k \pi}{2} ( k \in \mathbb{Z})\\ d)x=\dfrac{\pi}{4} + \dfrac{k  \pi}{2}  ( k \in \mathbb{Z})\\ e) x= -\dfrac{\pi}{6} + \dfrac{k 2 \pi}{3} ( k \in \mathbb{Z})\\ f) x= -\dfrac{\pi}{3} + \dfrac{k 2 \pi}{3} ( k \in \mathbb{Z})\\ g)\left[\begin{array}{cc} x = \dfrac{\arcsin \left(\dfrac{1}{5}\right)}{2} + k  \pi & ( k \in \mathbb{Z})  \\ x=\dfrac{\pi – \arcsin \left(\dfrac{1}{5}\right)}{2} + l  \pi & ( l \in \mathbb{Z}) \end{array} \right.\\ h) \left[\begin{array}{cc}  x=\dfrac{2\pi}{3} + k 2 \pi & ( k \in \mathbb{Z}) \\ x=\dfrac{-2\pi}{3} + l 2 \pi & ( l\in \mathbb{Z})\end{array} \right.$
    Lời giải và giải thích chi tiết:
    $a) \cos 2x – \sin x=0\\ \Leftrightarrow 1-2\sin^2x- \sin x=0\\ \Leftrightarrow 2\sin^2x+ \sin x-1=0\\ \Leftrightarrow 2\sin^2x+2 \sin x – \sin x -1=0\\ \Leftrightarrow 2\sin x( \sin x +1) -( \sin x + 1)=0\\ \Leftrightarrow (2\sin x – 1)( \sin x +1) =0\\ \Leftrightarrow \left[\begin{array}{l} 2\sin x – 1 =0\\ \sin x +1 =0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} 2\sin x  =1\\ \sin x  =-1\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \sin x  =\dfrac{1}{2}\\ \sin x  =-1\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \sin x  = \sin \left( \dfrac{\pi}{6} \right)\\ \sin x  =-1\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc}  x  =  \dfrac{\pi}{6} + k 2 \pi & ( k \in \mathbb{Z}) \\x  =  \dfrac{5\pi}{6} + l 2 \pi & ( l \in \mathbb{Z})  \\ x  =  -\dfrac{\pi}{2} + m 2 \pi & (m \in \mathbb{Z})  \end{array} \right.\\ b)\sin 2x+\sin x=0\\ \Leftrightarrow 2\sin x \cos x+\sin x=0\\ \Leftrightarrow \sin x (2\cos x+1)=0\\ \Leftrightarrow \left[\begin{array}{l} \sin x =0\\ 2\cos x+1=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \sin x =0\\ \cos x=-\dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \sin x =0\\ \cos x=\cos \left(\dfrac{2\pi}{3}\right)\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc} x = k \pi & ( k \in \mathbb{Z})  \\ x=\dfrac{2\pi}{3} + l 2 \pi & ( l \in \mathbb{Z}) \\ x=\dfrac{-2\pi}{3} + m 2 \pi & ( m\in \mathbb{Z})\end{array} \right.\\ c)\sin 2x=0\\ \Leftrightarrow 2x=k \pi ( k \in \mathbb{Z})\\ \Leftrightarrow x=\dfrac{k \pi}{2} ( k \in \mathbb{Z})\\ d)\cos 2x=0\\ \Leftrightarrow 2x=\dfrac{\pi}{2} + k  \pi  ( k \in \mathbb{Z})\\ \Leftrightarrow x=\dfrac{\pi}{4} + \dfrac{k  \pi}{2}  ( k \in \mathbb{Z})\\ e)\sin 3x=-1\\ \Leftrightarrow 3x= -\dfrac{\pi}{2} + k 2 \pi ( k \in \mathbb{Z})\\ \Leftrightarrow x= -\dfrac{\pi}{6} + \dfrac{k 2 \pi}{3} ( k \in \mathbb{Z})\\ f)\cos 3x=-1\\ \Leftrightarrow 3x= -\pi + k 2 \pi ( k \in \mathbb{Z})\\ \Leftrightarrow x= -\dfrac{\pi}{3} + \dfrac{k 2 \pi}{3} ( k \in \mathbb{Z})\\ g)\sin 2x=\dfrac{1}{5}\\ \Leftrightarrow \left[\begin{array}{cc} 2x = \arcsin \left(\dfrac{1}{5}\right) + k 2 \pi & ( k \in \mathbb{Z})  \\ 2x=\pi – \arcsin \left(\dfrac{1}{5}\right) + l 2 \pi & ( l \in \mathbb{Z}) \end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc} x = \dfrac{\arcsin \left(\dfrac{1}{5}\right)}{2} + k  \pi & ( k \in \mathbb{Z})  \\ x=\dfrac{\pi – \arcsin \left(\dfrac{1}{5}\right)}{2} + l  \pi & ( l \in \mathbb{Z}) \end{array} \right.\\ h)\cos x=-\dfrac{1}{2}\\ \Leftrightarrow \cos x=\cos \left(\dfrac{2\pi}{3}\right)\\ \Leftrightarrow \left[\begin{array}{cc}  x=\dfrac{2\pi}{3} + k 2 \pi & ( k \in \mathbb{Z}) \\ x=\dfrac{-2\pi}{3} + l 2 \pi & ( l\in \mathbb{Z})\end{array} \right.$

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

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