Question

очень срочно нужно желательно на листочке

Answer 1

$1)\; \; sinx-cosx=1+sinx\cdot cosx\\\\\star \; t=sinx-cosx\; ,\; \; t^2=\underline {sin^2x}-2sinx\cdot cosx+\underline {cos^2x}=1-2sinx\cdot cosx\; \to \\\\2sinx\cdot cosx=1-t^2\; ,\; \; sinx\cdot cosx=\frac{1-t^2}{2}\; \; \star \\\\t=1+\frac{1-t^2}{2}\; \; ,\; \; t-1=\frac{1-t^2}{2}\; \; ,\; \; 2(t-1)=1-t^2\; ,\; \; 2t-2=1-t^2\; ,\\\\t^2+2t-3=0\; ,\; \; t_1=1\; ,\; t_2=-3\; \; (teorema\; Vieta)\\\\a)\; \; sinx-cosx=1\; |:\sqrt2\\\\\frac{1}{\sqrt2}\cdot sinx-\frac{1}{\sqrt2}\cdot cosx=\frac{1}{\sqrt2}$

$(\frac{1}{\sqrt2})^2+(\frac{1}{\sqrt2})^2=1\; \; \to \; \; sin\frac{\pi}{4} =\frac{1}{\sqrt2}\; ,\; cos\frac{\pi}{4}=\frac{1}{\sqrt2}\; \; \; (sin^2a+cos^2a=1)\\\\cos\frac{\pi}{4}\cdot sinx-sin\frac{\pi}{4}\cdot cosx=\frac{1}{\sqrt2}\\\\sin(x-\frac{\pi}{4})=\frac{1}{\sqrt2}\\\\x-\frac{\pi }{4}=(-1)^{n}\cdot \frac{\pi}{4}+\pi n,\; n\in Z\\\\\underline {x=\frac{\pi }{4}+(-1)^{n}\cdot \frac{\pi }{4}+\pi n\; ,\; n\in Z}\\\\b)\; \; sinx-cosx=-3\; |:\sqrt2\\\\cos\frac{\pi}{4}\cdot sinx-sin\frac{\pi }{4}\cdot cosx=-\frac{3}{\sqrt2}$

$sin(x-\frac{\pi }{4})=-\frac{3}{\sqrt2}\; \; ,\; \; \; \; -\frac{3}{\sqrt2}<-1\; \; ,\; \; \; -1\leq sin(x-\frac{\pi}{4})\leq 1\; \; \Rightarrow \\\\x\in \varnothing \\\\Otvet:\; \; x=\frac{\pi}{4}+(-1)^{n}\cdot \frac{\pi}{4}+\pi n\; ,\; n\in Z\; .$

$2)\; \; 2sin^2x+5sinx\cdot cosx-5cos^2x=1\\\\2sin^2x+5sinx\cdot cosx-5cos^2x=sin^2x+cos^2x\\\\sin^2x+5sinx\cdot cosx-6cos^2x=0\; |:cos^2x\ne 0\\\\tg^2x+5tgx-6=0\; ,\; D=49\; ,\; \; (tgx)_1=-6\; ,\; (tgx)_2=1\\\\a)\; tgx=-6\; ,\; \underline {x=-arctg6+\pi n,\; n\in Z}\\\\b)\; tgx=1\; ,\; \underline {x=\frac{\pi}{4}+\pi k,\; k\in Z}$