Решить уравнение: [latex]2Cos(2x)=Sin^3(x)+Cos^3(x)[/latex]

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[latex] 2 \cos{ 2x } = \sin^3{x} + \cos^3{x} \ ; [/latex] [latex] 2 ( \cos^2{x} - \sin^2{x} ) = ( \sin{x} + \cos{x} ) ( \sin^2{x} - \sin{x} \cos{x} + \cos^2{x} ) \ ; [/latex] [latex] 2 ( \cos{x} + \sin{x} ) ( \cos{x} - \sin{x} ) = ( \sin{x} + \cos{x} ) ( 1 - \sin{x} \cos{x} ) \ ; [/latex] [latex] ( \cos{x} + \sin{x} ) ( 2 [ \cos{x} - \sin{x} ] + \sin{x} \cos{x} - 1 ) = 0 \ ; [/latex] [latex] ( \cos{x} + \cos{ [ \frac{ \pi }{2} - x ] } ) ( 2 [ \cos{x} + \sin{ ( - x ) } ] + \frac{1}{2} \sin{2x} - 1 ) = 0 \ ; [/latex] [latex] \cos{ \frac{ x + \pi/2 - x }{2} } \cos{ \frac{ x - [ \pi/2 - x ] }{2} } \cdot ( 2 [ \cos{x} + \cos{ ( \frac{ \pi }{2} - \{ - x \} ) } ] + \frac{1}{2} \cos{ ( \frac{ \pi }{2} - 2x ) } - 1 ) = 0 \ ; [/latex] [latex] \cos{ \frac{ \pi }{4} } \cos{ \frac{ x - \pi/2 + x }{2} } \cdot ( 2 [ \cos{x} + \cos{ ( x + \frac{ \pi }{2} ) } ] + \frac{1}{2} \cos{ ( 2 [ \frac{ \pi }{4} - x ] ) } - 1 ) = 0 \ ; [/latex] [latex] \cos{ \frac{ 2x - \pi/2 }{2} } \cdot ( 2 [ 2 \cos{ \frac{ x + x + \pi/2 }{2} } \cos{ \frac{ x - [ x + \pi/2 ] }{2} } ] + \frac{1}{2} [ 1 - 2 \sin^2{ ( \frac{ \pi }{4} - x ) } ] - 1 ) = 0 \ ; [/latex] [latex] \cos{ ( x - \frac{ \pi }{4} ) } \cdot ( 2 \cdot 2 \cos{ \frac{ 2x + \pi/2 }{2} } \cos{ \frac{ - \pi/2 }{2} } + \frac{1}{2} [ 1 - 2 \cos^2{ ( \frac{ \pi }{2} - \{ \frac{ \pi }{4} - x \} ) } ] - 1 ) = 0 \ ; [/latex] [latex] \cos{ ( x - \frac{ \pi }{4} ) } ( 4 \cos{ ( x + \frac{ \pi }{4} ) } \cos{ \frac{ \pi }{4} } + \frac{1}{2} [ 1 - 2 \cos^2{ ( \frac{ \pi }{2} - \frac{ \pi }{4} + x ) } ] - 1 ) = 0 \ ; [/latex] [latex] \cos{ ( x - \frac{ \pi }{4} ) } ( 4 \cos{ ( x + \frac{ \pi }{4} ) } \cdot \frac{ \sqrt{2} }{2} + \frac{1}{2} [ 1 - 2 \cos^2{ ( \frac{ \pi }{4} + x ) } ] - 1 ) = 0 \ ; [/latex] [latex] \cos{ ( x - \frac{ \pi }{4} ) } ( 2 \sqrt{2} \cos{ ( x + \frac{ \pi }{4} ) } + \frac{1}{2} - \cos^2{ ( x + \frac{ \pi }{4} ) } - 1 ) = 0 \ ; [/latex] [latex] \cos{ ( x - \frac{ \pi }{4} ) } ( 2 \sqrt{2} \cos{ ( x + \frac{ \pi }{4} ) } - \frac{1}{2} - \cos^2{ ( x + \frac{ \pi }{4} ) } ) = 0 \ ; [/latex] [latex] \cos{ ( x - \frac{ \pi }{4} ) } ( \cos^2{ ( x + \frac{ \pi }{4} ) } - 2 \sqrt{2} \cos{ ( x + \frac{ \pi }{4} ) } + \frac{1}{2} ) = 0 \ ; [/latex] [latex] \left\{\begin{array}{l} \cos{ ( x - \frac{ \pi }{4} ) } = 0 \ ; \\\\ \left|\begin{array}{l} D_1 = ( \sqrt{2} )^2 - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2} \ ; \\\\ \cos{ ( x + \frac{ \pi }{4} ) } = \sqrt{2} \pm \sqrt{ \frac{3}{2} } = \sqrt{2} - \sqrt{ \frac{3}{2} } \ ; \end{array}\right \end{array}\right [/latex] [latex] \left\{\begin{array}{l} x - \frac{ \pi }{4} = \frac{ \pi }{2} + \pi k , k \in Z \ ; \\ x + \frac{ \pi }{4} = \pm arccos{ ( \sqrt{2} - \sqrt{ \frac{3}{2} } ) } + 2 \pi n , n \in Z \ ; \end{array}\right [/latex] О т в е т : [latex] \left\{\begin{array}{l} x = \frac{ 3 \pi }{4} + \pi k , k \in Z \ ; \\ x = \pm arccos{ ( \sqrt{2} - \sqrt{ \frac{3}{2} } ) } - \frac{ \pi }{4} + 2 \pi n , n \in Z \ . \end{array}\right [/latex]