Đáp án:

Chọn C.

Giải thích các bước giải: \[\begin{array}{l} \left( {1 - \sin x} \right){\sin ^2}x - \left( {1 + \cos x} \right){\cos ^2}x = 0\\ \Leftrightarrow {\sin ^2}x - {\sin ^3}x - {\cos ^2}x - {\cos ^3}x = 0\\ \Leftrightarrow \left( {{{\sin }^2}x - {{\cos }^2}x} \right) - \left( {{{\sin }^3}x + {{\cos }^3}x} \right) = 0\\ \Leftrightarrow \left( {\sin x - \cos x} \right)\left( {\sin x + \cos x} \right) - \left( {\sin x + \cos x} \right)\left( {{{\sin }^2}x - \sin x\cos x + {{\cos }^2}x} \right) = 0\\ \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {\sin x - \cos x - 1 + \sin x\cos x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x + \cos x = 0\,\,\,\,\,\left( 1 \right)\\ \sin x - \cos x + \sin x\cos x - 1 = 0\,\,\,\,\,\left( 2 \right) \end{array} \right.\\ \left( 1 \right) \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = 0 \Leftrightarrow x + \frac{\pi }{4} = k\pi \Leftrightarrow x = - \frac{\pi }{4} + k\pi \,\,\,\,\left( {k \in Z} \right).\\ \Rightarrow y = - \frac{\pi }{4}.\\ \left( 2 \right) \Leftrightarrow \sin x - \cos x + \sin x\cos x - 1 = 0\\ Dat\,\,\,t = \sin x - \cos x\,\,\left( { - \sqrt 2 \le t \le \sqrt 2 } \right)\\ \Rightarrow {t^2} = 1 - 2\sin x\cos x \Rightarrow \sin x\cos x = \frac{{1 - {t^2}}}{2}\\ \Rightarrow \left( 2 \right) \Leftrightarrow t + \frac{{1 - {t^2}}}{2} - 1 = 0\\ \Leftrightarrow 2t + 1 - {t^2} - 2 = 0\\ \Leftrightarrow {t^2} - 2t + 1 = 0\\ \Leftrightarrow {\left( {t + 1} \right)^2} = 0\\ \Leftrightarrow t = - 1\\ \Leftrightarrow \sin x - \cos x = - 1\\ \Leftrightarrow \sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = - 1\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{4}} \right) = - \frac{1}{{\sqrt 2 }}\\ \Leftrightarrow \left[ \begin{array}{l} x - \frac{\pi }{4} = - \frac{\pi }{4} + k2\pi \\ x - \frac{\pi }{4} = \frac{{5\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k2\pi \\ x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right)\\ \Rightarrow \left\{ \begin{array}{l} \alpha = 0\\ \beta = \frac{{3\pi }}{2} \end{array} \right. \Rightarrow \alpha + \beta + y = 0 + \frac{{3\pi }}{2} - \frac{\pi }{4} = \frac{{5\pi }}{4}. \end{array}\]

$(1-\sin x){\sin}^2x-(1+\cos x){\cos}^2x=0$ $\Rightarrow{\sin}^2x-{\cos}^2x-{\sin}^3x-{\cos}^3x=0$ $\Rightarrow(\sin x-\cos x)(\sin x+\cos x)-(\sin x+\cos x)(1-\sin x\cos x)=0$ $\Rightarrow (\sin x+\cos x)(\sin x-\cos x-1+\sin x\cos x)=0$ $\Rightarrow \left[ \begin{array}{l}\sin x+\cos x=0(1)\\\sin x-\cos x-1+\sin x\cos x=0 (2) \end{array} \right .$ $(1)\Rightarrow \sqrt2\sin(x+\dfrac{\pi}{4})=0$ $\Rightarrow x+\dfrac{\pi}{4}=k\pi$ $\Rightarrow x=\dfrac{-\pi}{4}+k\pi$ $\Rightarrow\beta=\dfrac{-\pi}{4}$ $(2)\Rightarrow\sin x(1+\cos x)-(1+\cos x)=0 $ $\Rightarrow(\sin x-1)(\cos x+1)=0 $ $\Rightarrow \left[\begin{array}{l} \sin x=1 \\\cos x=-1 \end{array} \right .\Rightarrow \left[\begin{array}{l} x=\dfrac{\pi}{2}+k2\pi \\ x=\pi+k2\pi\end{array} \right .$ $\Rightarrow \left\{ \begin{array}{l} \alpha= \dfrac{\pi}{2}\\y=\pi\end{array} \right .$ $\Rightarrow\alpha+\beta+y=\dfrac{5\pi}{4}$.