Answers: 1
![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\\\\](/tpl/images/0177/4686/52a0a.png)
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Why can't you factor 2cosx^2+sinx-1=0 ?...
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