Question

Сложное тригонометрическое уравнение .

Answer 1

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Answer 2

$1-2sinx\cdot cosx-sinx-cosx=0\\\\1-2sinx\cdot cosx-(sinx+cosx)=0\\\\t=sinx+cosx\; ,\; \; t^2=1+2sinx\cdot cosx\; \to \; 2sinx\cdot cosx=t^2-1\\\\1-(t^2-1)-t=0\\\\1-t^2+1-t=0\\\\t^2+t-2=0\\\\t_1=-2,\; t_2=1\; (teorema\; Vieta)\\\\a)\; sinx+cosx=-2\\\\sinx+sin(\frac{\pi}{2}-x)=-2\\\\2sin\frac{\pi}{4}\cdot cos(x-\frac{\pi}{4})=-2\\\\\sqrt2\cdot cos(x- \frac{\pi }{4})=-2\\\\cos(x-\frac{\pi}{4})=-\sqrt2\; \; \; (-\sqrt2\approx -1,4\ \textless \ -1) \; \; \to \; \; net\; resenij\\\\b)\; \; sinx+cosx=1$

$\sqrt2\cdot cos(x-\frac{\pi}{4})=1$

$cos(x-\frac{\pi}{4})=\frac{\sqrt 2}{2}\\\\x-\frac{\pi}{4}=\pm \frac{\pi}{4}+2\pi n,\; n\in Z\\\\x=\frac{\pi}{4}\pm \frac{\pi}{4}+2\pi n= \left [ {{\frac{\pi}{2}+2\pi n,\; n\in Z} \atop {2\pi n,\; n\in Z}} \right. \; \; \; -\; \; otvet$