giải pt: 12/` 1+tanx=2\sqrt{2}sin(x+\frac{ π}{4}) `

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1. Đáp án:

   $1+tanx = 2\sqrt{2}sin(x+\frac{\pi}{4})$  $( ĐK : x\neq \frac{\pi}{2} + k\pi) $

   $⇔ 1 + \frac{sinx}{cosx} = 2(sinx + cosx )$

   $⇔ \frac{cosx+sinx}{cosx} – 2(sinx + cosx) = 0$

   $⇔ cosx + sinx – 2cosx(sinx + cosx) = 0 $

   $⇔ (sinx + cosx)(1 – 2cosx) = 0$

   $⇔$ \(\left[ \begin{array}{l}sinx+cosx=0\\1-2cosx=0\end{array} \right.\) 

   $⇔$\(\left[ \begin{array}{l}\sqrt{2} sin(x+\frac{\pi}{4})=0\\cosx=\frac{1}{2}\end{array} \right.\) 

   $⇔$ \(\left[ \begin{array}{l}x=-\frac{\pi}{4}+k\pi\\x=±\frac{\pi}{3}+k2\pi\end{array} \right.\)  $(TM)$

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