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Why can't you factor 2cosx^2+sinx-1=0 ?

Sagot :

[tex]2cos^2x+sinx-1=0\\\\2(1-sin^2x)+sinx-1=0\\\\2-2sin^2x+sinx-1=0\\\\-2sin^2x+sinx+1=0\\\\-2sin^2x+2sinx-sinx+1=0\\\\-2sinx(sinx-1)-1(sinx-1)=0\\\\(sinx-1)(-2sinx-1)=0\iff sinx-1=0\ or\ -2sinx-1=0\\\\sinx=1\ or\ -2sinx=1\\\\sinx=1\ or\ sinx=-\frac{1}{2}\\\\x=\frac{\pi}{2}+2k\pi\ or\ x=-\frac{\pi}{6}+2k\pi\ or\ x=\frac{7\pi}{6}+2k\pi\ where\ k\in\mathbb{Z}[/tex]
kate200468
[tex]2cosx^2+sinx-1=2(1-sin^2x)+snx-1=\\\\=2(1-sinx)(1+sinx)-(1-sinx)=(1-sinx)[2(1+sinx)-1]=\\\\=(1-sinx)(2+2sinx-1)=(1-sinx)(1+2sinx)\\\\2cosx^2+sinx-1=0\ \ \ \Leftrightarrow\ \ \ (1-sinx)(1+2sinx)=0\\\\1-sinx=0\ \ \ \ \ or\ \ \ \ \ 1+2sinx=0\\\\1)\ \ \ 1-sinx=0\ \ \ \Rightarrow\ \ \ sinx=1\ \ \ \Rightarrow\ \ \ x= \frac{ \pi }{2} +2k \pi ,\ \ \ k\in I\\\\[/tex]

[tex]2)\ \ \ 1+2sinx=0\ \ \ \ \ \ \Rightarrow\ \ \ sinx=- \frac{1}{2}\\\\ \Rightarrow\ \ \ x_1=( \pi + \frac{ \pi }{6} )+2k \pi ,\ \ \ \ \ \ x_2=( - \frac{ \pi }{6} )+2k \pi,\ \ \ \ \ \ k\in I\\\\.\ \ \ \ \ \ x_1=\frac{7 \pi }{6} +2k \pi ,\ \ \ \ \ \ \ \ \ \ \ \ x_2=-\frac{ \pi }{6} +2k \pi,\ \ \ \ \ \ \ \ \ k\in I\\\\Ans.\ x=-\frac{ \pi }{6} +2k \pi\ \ \ or\ \ \ x= \frac{ \pi }{2} +2k \pi\ \ \ or\ \ \ x=\frac{7 \pi }{6} +2k \pi,\ \ \ k\in I[/tex]
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