Giải phương trình $\frac{sin2x+cosx-\sqrt{3}(cos2x+sinx)}{2sin2x-\sqrt{3}}=1$
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$\frac{sin2x+cosx-\sqrt{3}(cos2x+sinx)}{2sin2x-\sqrt{3}}=1$
Started By nguyenvinhpthanh, 12-02-2016 - 12:25
#1
Posted 12-02-2016 - 12:25
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Posted 03-03-2016 - 21:54
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Giải phương trình $\frac{sin2x+cosx-\sqrt{3}(cos2x+sinx)}{2sin2x-\sqrt{3}}=1$
Điều kiện: $sin2x\neq \frac{\sqrt{3}}{2}$
Ta có:
$\frac{sin2x+cosx-\sqrt{3}(cos2x+sinx)}{2sin2x-\sqrt{3}}=1$
$\Leftrightarrow sin2x+cosx-\sqrt{3}cos2x-\sqrt{3}sinx=2sin2x-\sqrt{3}$
$\Leftrightarrow sin2x-cosx+\sqrt{3}cos2x+\sqrt{3}sinx-\sqrt{3}=0$
$\Leftrightarrow 2sinxcosx-cosx-2\sqrt{3}(sinx)^{2}+\sqrt{3}sinx=0$
$\Leftrightarrow 2sinx(cosx-\sqrt{3}sinx)-(cosx-\sqrt{3}sinx)=0$
$\Leftrightarrow (2sinx-1)(cosx-\sqrt{3}sinx)=0$
~~~~~~~~~~~~~~ Nếu bạn theo đuổi đam mê ....
thành công sẽ đuổi theo bạn!!!!!!!!!!!!!!!!!!!!!
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