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[hoctrocuanewton]

Giải phương trình sau :

$\frac{\sqrt{2}(2-\tan x)}{\sin (5x-\frac{\pi }{4})}=\frac{1+\tan x}{\sin x}$

[KySuBachKhoa]

$sinx\neq 0 , sin(5x-\frac{\pi }{4})\neq 0$

$\frac{\sqrt{2}\left ( 2- tanx \right )}{sin\left ( 5x-\frac{\pi }{4} \right )}= \frac{1}{sinx}+\frac{1}{cosx}=\frac{\sqrt{2}sin\left ( x+\frac{\pi }{4} \right )}{sinxcosx}$

$sin2x - sin^{2}x =\frac{1}{2}\left ( cos4x - cos6x \right )$

$2sin2x - cos2x +1 = cos4x - 4cos^{3}2x + 3cos2x$

$2sin2x + 1= cos4x + 4cos2xsin^{2}2x$

$2sin2x( 1- sin4x) + 1- cos4x =0$

$2sin2x( 1- sin4x + sin2x) = 0$

bạn tự giải nhé

Edited by KySuBachKhoa, 24-06-2015 - 00:56.