1-tanx/2)(1+sin) =1+tanx/2

Đáp án:

\[x = k2\pi \]

Giải thích các bước giải:

\[\begin{array}{l}
1 + \sin x = \frac{{1 + \tan \frac{x}{2}}}{{1 - \tan \frac{x}{2}}}\\
 \Leftrightarrow 1 + \sin x = \frac{{(c{\rm{os}}\frac{x}{2} + \sin \frac{x}{2})(c{\rm{os}}\frac{x}{2} - \sin \frac{x}{2})}}{{(c{\rm{os}}\frac{x}{2} - \sin \frac{x}{2})(c{\rm{os}}\frac{x}{2} - \sin \frac{x}{2})}} = \frac{{{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}}}{{{{(c{\rm{os}}\frac{x}{2} - \sin \frac{x}{2})}^2}}} = \frac{{{\rm{cosx}}}}{{1 - \sin x}}\\
 \Leftrightarrow 1 - {\sin ^2}x = c{\rm{osx}}\\
 \Leftrightarrow {\rm{1 - si}}{{\rm{n}}^2}x = 1 - 2{\sin ^2}\frac{x}{2}\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\sin x = \sqrt 2 \sin \frac{x}{2}}\\
{\sin x =  - \sqrt 2 \sin \frac{x}{2}}
\end{array}} \right. \Leftrightarrow x = k2\pi 
\end{array}\]

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