Toán Lớp 11: Giải phương trình lượng giác: `(sin^4 2x+cos^4 2x)/(tan(π/4-x).tan(π/4+x))=cos^4 4x`

Question

Toán Lớp 11:  Giải phương trình lượng giác: (sin^4 2x+cos^4 2x)/(tan(π/4-x).tan(π/4+x))=cos^4 4x

ngoclandiep · Answer

(sin^4 2x+cos^4 2x)/(tan(&pi;/4-x).tan(&pi;/4+x))= cos^4 x   $ Leftrightarrow  dfrac{ sin ^4 (2x)+  cos ^4(2x)}{ dfrac{  tan  Big( dfrac{ pi}{2}  Big) -  tan x}{1+  tan  Big( dfrac{ pi}{4} Big)  tan x}.   dfrac{ tan  Big( dfrac{ pi}{4} Big)+  tan x}{1- tan  Big( dfrac{ pi}{4} Big)  tan x}}=  cos ^4x$   $ Leftrightarrow  dfrac{ sin ^4 (2x)+  cos ^4(2x)}{ dfrac{1- tan x}{1+ tan x}.   dfrac{1+  tan x}{1- tan x}} =  cos ^4 x$   $ Leftrightarrow  sin ^4 (2x)+  cos ^4(2x) =  cos ^4 x$   $ Leftrightarrow (2  sin x.  cos x)^4 +  cos ^4 2x -   cos ^4 x =0$   $ Leftrightarrow 16  sin ^4 x  cos ^4 x +  cos ^8 x -   sin ^8 x -  cos ^4 x =0$   $ Leftrightarrow  cos ^4 x. (16  sin ^4x +  cos ^2 x-1- tan ^4 x. sin ^4 x) =0$   $ to x= dfrac{ pi}{2}+k pi$    Phương tr&igrave;nh sau vẫn chưa c&oacute; c&aacute;ch giải.

hauthanhyen98 · Answer

$ dfrac{ sin^42x+ cos^42x}{ tan  left( dfrac  pi 4-x right) tan  left( dfrac  pi 4+x right)}= cos^4 4x$   ĐKXĐ: $x ne  dfrac{ pi}4+ dfrac{k pi}2$   $&hArr;  sin^4 2x+ cos^4 2x= cos^4 x$ (Mẫu $= 1$)   $&hArr;(1- cos 4x)^2+(1+ cos 4x)^2=4 cos^4 4x$   $&hArr; 2 cos^44x- cos^24x-1=0$   $&hArr;  cos^24x-1=0$   $&hArr;  cos x= dfrac12$   $&hArr; x= pm  dfrac  pi3+k2 pi  (k in  mathbb{Z})$ (thỏa m&atilde;n)   Vậy $x= pm  dfrac  pi3+k2 pi  (k in  mathbb{Z})$