Giải phương trình lượng giác: cos2x+sin2x+9cosx-sinx-4=0.
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Pt đã cho tương đương: $2cos^2x-1+2sinxcosx+9cosx-sinx-4=0 \Leftrightarrow 2cos^2x+2sinxcosx+9cosx-sinx-5=0$
$\Leftrightarrow (2cosx-1)(cosx+sinx+5)=0\Leftrightarrow cosx=\frac{1}{2} hoặc cosx+sinx+5=0$
TH1: $cosx=\frac{1}{2}=cos(\pi/2)\Rightarrow x=\frac{\pm \pi }{3}+k2\pi $
TH2 $cosx+sinx+5=0\Leftrightarrow cosx+sinx=-5$  ( Vô nghiệm)
Chứng minh: $cosx+sinx\leq |cosx+sinx|\leq |cosx|+|sinx|\leq 2$ Do $cosx,sinx\in [-1;1]$
Vậy kết luận nghiệm.......
cosx sinx =< căn 2 nhé –  nguyenquangtruonghktcute 31-05-16 12:16 PM
sautatca là ặc của chú à –  nguyenquangtruonghktcute 31-05-16 12:15 PM
vote và tích nha –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 30-05-16 03:14 PM
$cos2x+sin2x+9cosx-sinx-4=0$
$\Leftrightarrow 2cos^2x-1+2sinx.cox+9cosx-sinx-4=0$
$(2cosx-1)(sinx+cosx+5)=0$
TH1:
$cosx=\frac{1}{2}\Leftrightarrow x=\frac{\pm \pi }{3}+k2\pi $
TH2
$cosx+sinx=-5$
vì $cosx+sinx=\sqrt{2}sin(x+\frac{\pi }{4})\Leftrightarrow -\sqrt{2}\leq sinx+cosx\leq \sqrt{2}$ nên TH2 vô nghiệm
nghiệm của pt là $x=\frac{\pm \pi }{3}+k2\pi $
này thì báo cáo chấp lượng thấp –  nguyenquangtruonghktcute 31-05-16 12:26 PM

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