[Toán 10] Giải giúp em một bài toán cực khó

H

hung_ils

[TEX]sinx+cosx-tan\frac{x}{2}=\sqrt{2}cosxsin(\frac{x}{2}+\frac{\pi }{4})=cosx(sin\frac{x}{2}+cos\frac{x}{2})\\\\\Rightarrow 2sin\frac{x}{2}cos\frac{x}{2}-\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=cosx(sin\frac{x}{2}+cos\frac{x}{2}-1)\Rightarrow \frac{2cos^{2}\frac{x}{2}-1}{cos\frac{x}{2}}*sin\frac{x}{2}=cosx(sin\frac{x}{2}+cos\frac{x}{2}-1)\\\\\Rightarrow \frac{cosx}{cos\frac{x}{2}}sin\frac{x}{2}=cosx(sin\frac{x}{2}+cos\frac{x}{2}-1)\\\\\Rightarrow \left\{\begin{matrix}cosx=0 & \\ sin\frac{x}{2}=cos\frac{x}{2}sin\frac{x}{2}+cos^{2}\frac{x}{2}-cos\frac{x}{2}=0 & \\ \end{matrix}\right.\Rightarrow \left\{\begin{matrix}cosx=0 & \\cos\frac{x}{2}(cos\frac{x}{2}-1)+sin\frac{x}{2}(cos\frac{x}{2}-1)=0 & \end{matrix}\right. [/TEX]
PS:Tất cả những chỗ x2 là x/2 đấy nhé. Chả hiểu sao nó lại như vậy.........
 
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B

bin96vn

Bạn có thể làm lại được không, mình nhìn không hiểu cái gì cả! Thanks trước!:p
 
H

hung_ils

[TEX]sinx+cosx-tan\frac{x}{2}=cosx(sin(\frac{x}{2})+cos(\frac{x}{2}))\\\\\Rightarrow 2sin(\frac{x}{2})cos(\frac{x}{2})-\frac{sin(\frac{x}{2})}{cos(\frac{x}{2})}=cosx(sin(\frac{x}{2})+cos(\frac{x}{2})-1)\\\\\Rightarrow \frac{sin(\frac{x}{2})*(2cos^{2}(\frac{x}{2})-1)}{cos(\frac{x}{2})}=cosx(sin(\frac{x}{2})+cos(\frac{x}{2})-1)\\\\\Rightarrow tan(\frac{x}{2})cosx=cosx(sin(\frac{x}{2})+cos(\frac{x}{2})-1)\\\\\Rightarrow \left\{\begin{matrix}cosx=0 & \\ sin(\frac{x}{2})=cos(\frac{x}{2})sin(\frac{x}{2})+cos^{2}(\frac{x}{2})-cos(\frac{x}{2})\Rightarrow cos(\frac{x}{2})(cos(\frac{x}{2})-1)+sin(\frac{x}{2})(cos(\frac{x}{2})-1)=0 & \end{matrix}\right.[/TEX]
PS:Hóa ra là dấu ngoặc..........
 
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