Toán Lớp 11: Giải pt sau lsinx -cosx l +4. sin2x=1

Question

Toán Lớp 11:  Giải pt sau   lsinx -cosx l +4. sin2x=1

dantam45 · Answer

$ t =  left| { sin x -  cos x}  right| left( {0  le t  le  sqrt 2 }  right)     Rightarrow {t^2} = { left( { sin x -  cos x}  right)^2} = { sin ^2}x + { cos ^2}x - 2 sin x cos x = 1 - 2 sin x cos x     Leftrightarrow 2 sin x cos x = 1 - {t^2}  Leftrightarrow  sin 2x = 1 - {t^2}   PT  Leftrightarrow t + 4 left( {1 - {t^2}}  right) = 1     Leftrightarrow 4{t^2} - t - 3 = 0     Leftrightarrow  left[  begin{array}{l} t = 1(TM)   t =  -  dfrac{3}{4}(L)  end{array}  right.  Rightarrow  left| { sin x -  cos x}  right| = 1     Rightarrow  left| { sqrt 2  sin  left( {x -  dfrac{ pi }{4}}  right)}  right| = 1     Leftrightarrow  left[  begin{array}{l}  sqrt 2  sin  left( {x -  dfrac{ pi }{4}}  right) = 1    sqrt 2  sin  left( {x -  dfrac{ pi }{4}}  right) =  - 1  end{array}  right.  Leftrightarrow  left[  begin{array}{l}  sin  left( {x -  dfrac{ pi }{4}}  right) =  dfrac{{ sqrt 2 }}{2}    sin  left( {x -  dfrac{ pi }{4}}  right) =  -  dfrac{{ sqrt 2 }}{2}  end{array}  right.     Leftrightarrow  left[  begin{array}{l} x -  dfrac{ pi }{4} =  dfrac{ pi }{4} + k2 pi    x -  dfrac{ pi }{4} =  dfrac{{3 pi }}{4} + k2 pi    x -  dfrac{ pi }{4} =  -  dfrac{ pi }{4} + k2 pi    x -  dfrac{ pi }{4} =  dfrac{{5 pi }}{4} + k2 pi   end{array}  right.  Leftrightarrow  left[  begin{array}{l} x =  dfrac{ pi }{2} + k2 pi    x =  pi  + k2 pi    x = k2 pi    x =  dfrac{{3 pi }}{2} + k2 pi   end{array}  right. left( {k  in  mathbb{Z}}  right) $