tính tổng nghiệm thuộc đoạn

Question

tính tổng nghiệm thuộc đoạn [0;2013] của phương trình

$2sin^{2}(x-\frac{\pi }{4})=2sin^{2}x-tanx$

score 1 · Answer 1

$2sin^{2}(x-\frac{\pi }{4})=2sin^{2}x-tanx\Leftrightarrow 2[\frac{\sqrt{2}}{2}(sinx-sosx)]^{2}=2sin^{2}x-\frac{sinx}{cosx}$

$\Leftrightarrow (sinx-cosx)^{2}=sinx\frac{(2sinx.cosx-1)}{cosx}$

$\Leftrightarrow (sinx-cosx)^{2}=-sinx.\frac{(sinx-cosx)^{2}}{cosx}$

$\Leftrightarrow (sinx-cosx)^{2}.(1+tanx)=0$

$\Leftrightarrow$

$\left [ \begin{matrix} sinx-cosx=0 \\ 1+tanx=0 \end{matrix} \right.$

$\Leftrightarrow$

$\left [ \begin{matrix} sin(x-\frac{\pi }{4})=0 \\ tanx=-1 \end{matrix} \right.$

$\Leftrightarrow$

$\left [ \begin{matrix} x=\frac{\pi }{4}+k\pi \\ x=-\frac{\pi }{4}+k\pi \end{matrix} \right.$

$\Leftrightarrow x=\frac{\pi }{4}+k\pi ,k\in Z$

ma

$x\in [1;2013]$

$\Leftrightarrow x=\frac{\pi}{4}+k\pi,k\in [1;640]$ ,

$k\in Z$

Vay tong cac nghiem cua pt la:

$x_{1}+x_{2}+x_{3}+..+x_{640}=\frac{\pi}{4}+\pi+\frac{\pi}{4}+2\pi+\frac{\pi}{4}+3\pi+...+\frac{\pi}{4}+640\pi$

$=640.\frac{\pi}{4}+\pi(1+2+3+...+640)=160\pi+640.(\frac{640+1}{2})$

$=205280\pi$

1 Đáp án