Toán Lớp 11: (2sinx-1) (2cos2x+2sinx+3)+1=4sin^2x

Question

Toán Lớp 11:  (2sinx-1) (2cos2x+2sinx+3)+1=4sin^2x

haiyemy · Answer

$(2 sin x-1)(2 cos2x+2 sin x+3)+1=4 sin^2x$   $ to (2 sin x-1)(2 cos 2x+2 sin x+3)=4 sin^2x-1=(2 sin x-1)(2 sin x+1)$   $ to (2 sin x-1)(2 cos2x+2 sin x+3-2 sin x-1)=0$   $ to (2 sin x-1)(2 cos2x+2)=0$   $ to  left[  begin{array}{l} sin x= dfrac{1}{2}   cos2x=-1  end{array}  right.$   $ to  left[  begin{array}{l}x= dfrac{ pi}{6}+k2 pi    x= dfrac{5 pi}{6}+k2 pi    2x= pi+k2 pi  end{array}  right.$   $ to  left[  begin{array}{l}x= dfrac{ pi}{6}+k2 pi    x= dfrac{5 pi}{6}+k2 pi   x= dfrac{ pi}{2}+k pi  end{array}  right.$

dananhnhu2k4 · Answer

$  left( {2 sin x - 1}  right) left( {2 cos 2x + 2 sin x + 3}  right) + 1 = 4{ sin ^2}x     Leftrightarrow  left( {2 sin x - 1}  right) left( {2 cos 2x + 2 sin x + 3}  right) -  left( {4{{ sin }^2}x - 1}  right) = 0     Leftrightarrow  left( {2 sin x - 1}  right) left( {2 cos 2x + 2 sin x + 3}  right) -  left( {2 sin x - 1}  right) left( {2 sin x + 1}  right) = 0     Leftrightarrow  left( {2 sin x - 1}  right) left( {2 cos 2x + 2 sin x + 3 - 2 sin x - 1}  right) = 0     Leftrightarrow  left( {2 sin x - 1}  right) left( {2 cos 2x + 2}  right) = 0     Leftrightarrow  left[  begin{array}{l}  sin x =  dfrac{1}{2}    cos 2x =  - 1  end{array}  right.  Leftrightarrow  left[  begin{array}{l} x =  dfrac{ pi }{6} + k2 pi    x =  dfrac{{5 pi }}{6} + k2 pi    2x =  pi  + k2 pi   end{array}  right.     Leftrightarrow  left[  begin{array}{l} x =  dfrac{ pi }{6} + k2 pi    x =  dfrac{{5 pi }}{6} + k2 pi    x =  dfrac{ pi }{2} + k pi   end{array}  right. left( {k  in  mathbb{Z}}  right) $