Đáp án:

$x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\left( {k \in Z} \right)$

Giải thích các bước giải:

 Ta có:

$\begin{array}{l}
\dfrac{{{{\sin }^4}\dfrac{x}{2} + {{\cos }^4}\dfrac{x}{2}}}{{1 - \sin x}} - {\tan ^2}x\sin x = \dfrac{{1 + \sin x}}{2} + {\tan ^2}x\left( {DK:\cos x \ne 0} \right)\\
 \Leftrightarrow \dfrac{{{{\left( {{{\sin }^2}\dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2}} \right)}^2} - 2{{\sin }^2}\dfrac{x}{2}{{\cos }^2}\dfrac{x}{2}}}{{1 - \sin x}} - {\tan ^2}x\sin x - \dfrac{{1 + \sin x}}{2} - {\tan ^2}x = 0\\
 \Leftrightarrow \dfrac{{1 - \dfrac{1}{2}{{\sin }^2}x}}{{1 - \sin x}} - \dfrac{{1 + \sin x}}{2} - {\tan ^2}x\left( {\sin x + 1} \right) = 0\\
 \Leftrightarrow \dfrac{{2\left( {1 - \dfrac{1}{2}{{\sin }^2}x} \right) - \left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}{{2\left( {1 - \sin x} \right)}} - {\tan ^2}x\left( {\sin x + 1} \right) = 0\\
 \Leftrightarrow \dfrac{{2 - {{\sin }^2}x - 1 + {{\sin }^2}x}}{{2\left( {1 - \sin x} \right)}} - \dfrac{{{{\sin }^2}x\left( {\sin x + 1} \right)}}{{{{\cos }^2}x}} = 0\\
 \Leftrightarrow \dfrac{1}{{2\left( {1 - \sin x} \right)}} - \dfrac{{{{\sin }^2}x\left( {\sin x + 1} \right)}}{{{{\cos }^2}x}} = 0\\
 \Leftrightarrow \dfrac{1}{{2\left( {1 - \sin x} \right)}} - \dfrac{{{{\sin }^2}x\left( {\sin x + 1} \right)}}{{1 - {{\sin }^2}x}} = 0\\
 \Leftrightarrow \dfrac{1}{{2\left( {1 - \sin x} \right)}} - \dfrac{{{{\sin }^2}x\left( {\sin x + 1} \right)}}{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}} = 0\\
 \Leftrightarrow \dfrac{1}{{2\left( {1 - \sin x} \right)}} - \dfrac{{{{\sin }^2}x}}{{1 - \sin x}} = 0\\
 \Leftrightarrow \dfrac{{1 - 2{{\sin }^2}x}}{{2\left( {1 - \sin x} \right)}} = 0\\
 \Rightarrow 1 - 2{\sin ^2}x = 0\\
 \Leftrightarrow \cos 2x = 0\\
 \Leftrightarrow 2x = \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\left( {k \in Z} \right)
\end{array}$

Vậy phương trình có họ nghiệm là: $x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\left( {k \in Z} \right)$