Question

Помогите, прошу :с
√2 sin(2x+π/4)+√2cosx=sin2x-1

Answer 1

Ответ:

$\frac{\pi }{2} +\pi n,~n\in\mathbb {Z}; \pm\frac{3\pi }{4} +2\pi k,~k\in\mathbb {Z}$

Пошаговое объяснение:

$\sqrt{2} sin(2x+\frac{\pi }{4} )+\sqrt{2} cosx= sin2x-1.$

Применим формулу:

$sin(\alpha +\beta )= sin\alpha cos\beta +cos\alpha sin\beta$

$\sqrt{2} ( sin2x*cos\frac{\pi }{4} +cos2x*sin\frac{\pi }{4} ) +\sqrt{2} cosx=sin2x-1;\\\sqrt{2}(sin2x* \frac{\sqrt{2} }{2} +cos2x*\frac{\sqrt{2} }{2}) +\sqrt{2} cosx=sin2x-1;\\ sin2x+cos2x +\sqrt{2} cosx-sin2x+1=0;\\cos2x+\sqrt{2} cosx+1=0;$

Применим формулу косинуса двойного угла:

$cos2x= 2cos^{2} x-1.$

$2cos^{2} x-1+\sqrt{2}cosx+1=0;\\ 2cos^{2} x+\sqrt{2} cosx=0;\\cosx( 2cosx+\sqrt{2} )=0;\\\left [\begin{array}{lcl} {{cosx=0,} \\ {2cosx+\sqrt{2} =0;}} \end{array} \right.\Leftrightarrow\left [ \begin{array}{lcl} {{cosx=0,} \\ {cosx=-\frac{\sqrt{2} }{2} };} \end{array} \right.\Leftrightarrow \left [ \begin{array}{lcl} {{x=\frac{\pi }{2}+\pi n, ~n\in\mathbb {Z},} \\\\ {x=\pm\frac{3\pi }{4}+2\pi k~k\in\mathbb {Z}. }} \end{array} \right.$