Đáp án:
$x = \left\{\dfrac{\pi}{3};\dfrac{3\pi}{4};\dfrac{7\pi}{4}\right\}$
Giải thích các bước giải:
$1 + tanx = 2\sqrt{2}.sin(x+\dfrac{\pi}{4})\,\,\,\,\, (ĐK: \, x \ne \dfrac{\pi}{2} + k\pi)\\ \Leftrightarrow \dfrac{cosx+six}{cosx} = 2\sqrt{2}.sin(x+\dfrac{\pi}{4})\\ \Leftrightarrow cosx + sinx = 2\sqrt{2}.cosx.sin(x+\dfrac{\pi}{4})\\ \Leftrightarrow \sqrt{2}sin(x+\dfrac{\pi}{4}) = 2\sqrt{2}cosx.sin(x+\dfrac{\pi}{4})\\ \Leftrightarrow sin(x+\dfrac{\pi}{4})(2cosx -1)=0\\ \Leftrightarrow \left[\begin{array}{l}sin(x+\dfrac{\pi}{4}) = 0\\cosx = \dfrac{1}{2}\end{array}\right.\\ \Leftrightarrow \left[ \begin{array}{l}x = -\dfrac{\pi}{4} + k\pi\\x=\dfrac{\pi}{3}+k2\pi\\x=-\dfrac{\pi}{3} + k2\pi\end{array} \right.\,\,\,(k\in \Bbb Z)$
Do $x \in (0;2\pi)$
nên $x = \left\{\dfrac{\pi}{3};\dfrac{3\pi}{4};\dfrac{7\pi}{4}\right\}$
