Toán Lớp 11: Giải pt lượng giác sau; Sinx + cosx =-1 Sin^2x -4sinx +3 =0

Question

Toán Lớp 11:  Giải pt lượng giác sau; Sinx + cosx =-1 Sin^2x -4sinx +3 =0

diepbich93 · Answer

sinx + cosx =-1 ( &radic;a&sup2;+b&sup2; = &radic;1&sup2;+1&sup2; = &radic;2 )   &hArr; &radic;2 ( sinx. 1/&radic;2 + cosx. 1/&radic;2 ) =-1   &hArr; &radic;2 ( sinx. cos &pi;/4 + cosx. sin &pi;/4 ) =-1   &hArr; &radic;2 sin ( x+ &pi;/4 ) = -1   &hArr; sin( x + &pi;/4 ) = -1/&radic;2   &hArr; sin ( x+&pi;/4) = sin ( -&pi;/4)   &hArr; x + &pi;/4 = -&pi;/4 + k2&pi;         x + &pi;/4= &pi; - -  &pi;/4 +k2&pi;   &hArr; x + &pi;/4 = -&pi;/4 +k2&pi;        x + &pi;/4 = 5&pi;/4 + k2&pi;   &hArr; x = -&pi;/4 - &pi;/4 + k2&pi;        x= 5&pi;/4 - &pi;/4 + k2&pi;   &hArr;x= -&pi;/2 + k2&pi;       x= &pi; + k2&pi;  ( k&isin;Z)   b/  sin&sup2;x - 4sinx + 3 = 0   * Đặt t = sinx ( -1&le; x &le; 1 ) c&oacute; :   t&sup2; - 4t + 3 = 0    $ left  { {{y=3} ( loại) atop {x=1}( nhận)}  right.$   => sinx = 1    &hArr;      x = &pi;/2 + k2&pi;, k &isin; Z    Vậy pt c&oacute; nghiệm l&agrave; : x = &pi;/2 + k2&pi; với k &isin; Z

thanhtung · Answer

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