Toán Lớp 11: cos2x+2sinx+m-1=0
a)giải pt với m=0
b)tìm m để pt có nghiệm
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Toán Lớp 11: cos2x+2sinx+m-1=0
a)giải pt với m=0
b)tìm m để pt có nghiệm
Comments ( 2 )
a)m = 0\\
\Leftrightarrow \cos 2x + 2\sin x – 1 = 0\\
\Leftrightarrow 1 – 2{\sin ^2}x + 2\sin x – 1 = 0\\
\Leftrightarrow – 2{\sin ^2}x + 2\sin x = 0\\
\Leftrightarrow – 2\sin x\left( {\sin x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
Vậy\,x = k\pi ;x = \dfrac{\pi }{2} + k2\pi \\
b)\cos 2x + 2\sin x + m – 1 = 0\\
\Leftrightarrow 1 – 2{\sin ^2}x + 2\sin x + m – 1 = 0\\
\Leftrightarrow 2{\sin ^2}x – 2\sin x – m = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\Delta ‘ \ge 0\\
– 1 \le \sin x \le 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
1 – 2.\left( { – m} \right) \ge 0\\
0 \le \dfrac{{ – m}}{2} < 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ge – \dfrac{1}{2}\\
m \le 0
\end{array} \right.\\
Vậy\, – \dfrac{1}{2} \le m \le 0
\end{array}$
<=> -2sin$^{2}$ x + 2sinx = 0
<=> 2sinx(1-sinx)=0
<=> \(\left[ \begin{array}{l}sinx=0\\1-sinx=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=k\pi\\x = \pi/2 + k2\pi \end{array} \right.\)
b) cos2x + 2sinx – 1=-m
<=> -sin$^{2}$ x + 2sinx =-m
<=>sin$^{2}$ x – 2sinx =m
ta có m = sin$^{2}$ x – 2sinx
=>m= $(sinx-1)^{2}$ -1
ta có $(sinx-1)^{2}$ -1 $\geq$ -1
=> m $\geq$ -1 thì pt có nghiệm