Giải phương trình: si{n^3}x + co{s^3}x = 2 - si{n^4}x\left( * \right).
Giải:
\begin{array}{l} \left( * \right) \Leftrightarrow cosx.co{s^2}x + \left( {{{\sin }^4}x + {{\sin }^3}x - 2} \right) = 0\\ \Leftrightarrow cos\,x\left( {1 - {{\sin }^2}x} \right) + \left( {\sin \,x - 1} \right)\left( {{{\sin }^3}x + 2{{\sin }^2}x + 2\sin x + 2} \right) = 0\\ \Leftrightarrow cos\,x\left( {1 - \sin \,x} \right)\left( {1 + \sin \,x} \right) - \left( {{{\sin }^3}x + 2{{\sin }^2}x + 2\sin x + 2} \right) = 0\\ \Leftrightarrow \left( {1 - \sin \,x} \right)\left[ {cos\,x\left( {1 - \sin \,x} \right) - \left( {{{\sin }^3}x + 2{{\sin }^2}x + 2\sin x + 2} \right)} \right] = 0\\ \Leftrightarrow \left( {1 - \sin \,x} \right)\left[ {cos\,x\left( {1 + \sin \,x} \right) - {{\sin }^2}x\left( {1 + \sin \,x} \right) - 2\left( {1 + \sin \,x} \right) - {{\sin }^2}x} \right] = 0\\ \Leftrightarrow \left( {1 - \sin \,x} \right)\left[ {\left( {1 + \sin \,x} \right)\left( {cos\,x - {{\sin }^2}x - 2} \right) - {{\sin }^2}x} \right] = 0\\ Do\,\left\{ \begin{array}{l} \left( {1 + \sin \,x} \right)\left( {cos\,x - {{\sin }^2}x - 2} \right) - {\sin ^2}x = \left( {1 + \sin \,x} \right)\left( {co{s^2}x + cos\,x - 3} \right) - {\sin ^2}x\\ co{s^2}x + cos\,x - 3 < 0\& 1 + \sin \,x < 0 \Rightarrow \left( {1 + \sin \,x} \right)\left( {co{s^2}x + cos\,x - 3} \right)\\ - {\sin ^2}x \le 0 \end{array} \right.\\ \Rightarrow \left( {1 + \sin \,x} \right)\left( {cos\,x - {{\sin }^2}x - 2} \right) - {\sin ^2}x \Rightarrow 1 - \sin \,x = 0 \Leftrightarrow \sin \,x = 1 \Leftrightarrow x = \frac{\pi }{2} + k2\pi \left( {k \in Z} \right) \end{array}
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