Giải phương trình: 1 + tanx = sinx + cosx.

ĐKXĐ: cosx ≠ 0 \( \Leftrightarrow x \ne \frac{\pi }{2} + k\pi \left( {k \in \mathbb{Z}} \right)\)

Ta có:

1 + tanx = sinx + cosx

\( \Leftrightarrow 1 + \tan x = \sin x + \cos x \Leftrightarrow 1 + \frac{{\sin x}}{{\cos x}} = \sin x + \cos x\)

\( \Leftrightarrow \frac{{\cos x + \sin x}}{{\cos x}} = \sin x + \cos x \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {\frac{1}{{\cos x}} - 1} \right) = 0\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin x + \cos x = 0}\\{\frac{1}{{\cos x}} = 1}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = 0}\\{\cos x = 1}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin \left( {x + \frac{\pi }{4}} \right) = 0}\\{\cos x = 1}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x + \frac{\pi }{4} = k\pi }\\{x = k2\pi }\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = - \frac{\pi }{4} + k\pi }\\{x = k2\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\).