Toán Lớp 11: sin2x + cosx =( 2sinx +1)( 3cosx -1) ????

Question

Toán Lớp 11:  sin2x + cosx =( 2sinx +1)( 3cosx -1) ????

huongtructram · Answer

$ begin{array}{l}  sin 2x +  cos x =  left( {2 sin x + 1}  right) left( {3 cos x - 1}  right)     Leftrightarrow  sin 2x +  cos x = 6 sin x cos x - 2 sin x + 3 cos x - 1     Leftrightarrow  sin 2x +  cos x = 3 sin 2x - 2 sin x + 3 cos x - 1     Leftrightarrow 1 - 2 sin 2x = 2 cos x - 2 sin x     Leftrightarrow 2 - 2 sin 2x = 2 left( { cos x -  sin x}  right) + 1     Leftrightarrow 2 left( {1 -  sin 2x}  right) = 2 left( { cos x -  sin x}  right) + 1     Leftrightarrow 2 left( {{{ sin }^2}x + {{ cos }^2}x - 2 sin x cos x}  right) = 2 left( { cos x -  sin x}  right) + 1     Leftrightarrow 2{ left( { cos x -  sin x}  right)^2} = 2 left( { cos x -  sin x}  right) + 1     Leftrightarrow 2{t^2} - 2t - 1 = 0 left( {t =  cos x -  sin x, left| t  right|  le  sqrt 2 }  right)     Leftrightarrow  left[  begin{array}{l} t =  dfrac{{1 +  sqrt 3 }}{2}   t =  dfrac{{1 -  sqrt 3 }}{2}  end{array}  right.  Leftrightarrow  left[  begin{array}{l}  cos x -  sin x =  dfrac{{1 +  sqrt 3 }}{2}    cos x -  sin x =  dfrac{{1 -  sqrt 3 }}{2}  end{array}  right.     Leftrightarrow  left[  begin{array}{l}  -  sqrt 2  sin  left( {x -  dfrac{ pi }{4}}  right) =  dfrac{{1 +  sqrt 3 }}{2}    -  sqrt 2  sin  left( {x -  dfrac{ pi }{4}}  right) =  dfrac{{1 -  sqrt 3 }}{2}  end{array}  right.     Leftrightarrow  left[  begin{array}{l}  sin  left( {x -  dfrac{ pi }{4}}  right) =  -  dfrac{{ sqrt 6  +  sqrt 2 }}{4}    sin  left( {x -  dfrac{ pi }{4}}  right) =  dfrac{{ sqrt 6  -  sqrt 2 }}{4}  end{array}  right.  end{array}$     Phương tr&igrave;nh cơ bản, tự t&iacute;nh nốt nh&eacute;.