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

Question

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

tuyetanhthu · Answer

Giải đáp+ Lời giải và giải thích chi tiết: <=> (2sinx+1).(2sin2x-1)=(2sinx+1).(2sinx-1) <=> [ sinx= (-1/2); sin2x=sinx] <=> x= -π/6+ k2π <=> [x=7π/6+K2π; 2x=x+2π] <=> 2x= r-x+2π <=> x= -π/6+k2π <=> [ x= 7π/+K2π; x= k2π] Vậy x= π/3+ k2π/3

tonhu12 · Answer

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