Giúp em với! a) Sinx+2cosx +3tan(x/2) =4 b) 1+ 2cos^2(3x/5) =3cos(4x/5) c) 3cos4x – 2cos^2(3x) =1

0 bình luận về “Giúp em với! a) Sinx+2cosx +3tan(x/2) =4 b) 1+ 2cos^2(3x/5) =3cos(4x/5) c) 3cos4x – 2cos^2(3x) =1”

1. \[\begin{array}{l}
   a)\,\,\sin \,x + 2\cos x + 3\tan \,\frac{x}{2} = 4\,\,\left( * \right)\\
   DK:\,\,\,\cos \left( {\frac{x}{2}} \right) \ne 0\\
   Dat\,\,\tan \,\frac{x}{2} = t \Rightarrow \sin t = \frac{{2t}}{{1 + {t^2}}};\,\,\,\,\,\cos t = \frac{{1 – {t^2}}}{{1 + {t^2}}}\\
   \Rightarrow \left( * \right) \Leftrightarrow \frac{{2t}}{{1 + {t^2}}} + 2.\frac{{1 – {t^2}}}{{1 + {t^2}}} + 3t = 4\\
   \Leftrightarrow 2t + 2\left( {1 – {t^2}} \right) + 3t\left( {1 + {t^2}} \right) = 4\left( {1 + {t^2}} \right)\\
   \Leftrightarrow 2t + 2 – 2t + 3t + 3{t^3} – 4 – 4{t^2} = 0\\
   \Leftrightarrow 3{t^3} – 4{t^2} + 3t – 2 = 0\\
   \Leftrightarrow \left( {t – 1} \right)\left( {3{t^2} – t + 2} \right) = 0\\
   \Leftrightarrow t = 1\\
   \Leftrightarrow \tan \frac{x}{2} = 1.\\
   b)\,\,1 + 2{\cos ^2}\left( {\frac{{3x}}{5}} \right) = 3\cos \left( {\frac{{4x}}{5}} \right)\\
   \Leftrightarrow 1 + \cos \frac{{6x}}{5} + 1 = 3\cos \frac{{4x}}{5}\\
   \Leftrightarrow \cos \frac{{6x}}{5} + 2 = 3\cos \frac{{4x}}{5}\\
   \Leftrightarrow 4{\cos ^3}\frac{{2x}}{5} – 3\cos \frac{{2x}}{5} + 2 = 6{\cos ^2}\frac{{2x}}{5} – 3\\
   \Leftrightarrow 4{\cos ^3}\frac{{2x}}{5} – 3\cos \frac{{2x}}{5} – 6{\cos ^2}\frac{{2x}}{5} + 5 = 0\\
   \Leftrightarrow \left( {\cos \frac{{2x}}{5} – 1} \right)\left( {4{{\cos }^2}\frac{{2x}}{5} – 2\cos \frac{{2x}}{5} – 5} \right) = 0
   \end{array}\]
   Câu c em làm tương tự nhé!!!

   Bình luận