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September 2021 1 7 Report

Решите неравенство
[tex] \sin^2 x-6\sin x\cos x+5\cos^2 x\ \textgreater \ 0;\\
1+2\sin x \geq 4\sin x \cos x+2\cos x[/tex]

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$1)\quad sin^2x-6sinx\cdot cosx+5cos^2x\ \textgreater \ 0\; |:cos^2x\ \textgreater \ 0,cosx\ne 0\\\\tg^2x-6tgx+5\ \textgreater \ 0\\\\(tgx)_1=1\; \; ili\; \; (tgx)_2=5\; ,\qquad +++(1)---(5)+++\\\\tgx\in (-\infty ,1)\cup (5,+\infty )\\\\ \left [ {{tgx\ \textless \ 1} \atop {tgx\ \textgreater \ 5}} \right. \; \left [ {{x_1\in (-\frac{\pi}{2}+\pi n,\; \frac{\pi}{4}+\pi n)} \atop {x_2\in (arctg5+\pi n,\; \frac{\pi}{2}+\pi n)}} \right. \\\\x\in (-\frac{\pi}{2}+\pi n,\; \frac{\pi}{4}+\pi n)\cup (arctg5+\pi n,\frac{\pi}{2}+\pi n)\; ,n\in Z$

$2)\quad 1+2sinx \geq 4sinx\cdot cosx+2cosx\\\\2(sinx-cosx)-4sinx\cdot cosx+1 \geq 0\\\\t=sinx-cosx\; \; \to \; \; t^2=(sinx-cosx)^2=1-2sinx\cdot cosx\; \to \\\\2sinx\cdot cosx=1-t^2\; \; \to \\\\2t-2(1-t^2)+1 \geq 0\\\\2t^2+2t-1 \geq 0\\\\D/4=1+2=3\; ,\; \; t_{1,2}= \frac{-1\pm \sqrt3}{2}\\\\+++( \frac{-1-\sqrt3}{2} )---( \frac{-1+\sqrt3}{2} )+++ \\\\t\in (-\infty , \frac{-1-\sqrt3}{2} \, ]\cup[\, \frac{-1+\sqrt3}{2} ,+\infty )\\\\a)\; \; sinx-cosx \leq \frac{-1-\sqrt3}{2} |:\sqrt2$

$\frac{1}{\sqrt2}sinx- \frac{1}{\sqrt2} cosx \leq \frac{-1-\sqrt3}{2\sqrt2} \\\\cos\frac{\pi}{4}\cdot sinx-sin\frac{\pi}{4}\cdot cosx \leq \frac{-1-\sqrt3}{2\sqrt2} \\\\sin(x-\frac{\pi}{4} )\leq \frac{-1-\sqrt3}{2\sqrt2} \; ,\; \; \frac{-1-\sqrt3}{2\sqrt2} \approx -0,97$

$arcsin \frac{-1-\sqrt3}{2\sqrt2} +2\pi n\leq x-\frac{\pi}{4} \leq \pi +arcsin \frac{-1-\sqrt3}{2\sqrt2} +2\pi n\; ,\; n\in Z$

$\frac{\pi}{4}+arcsin \frac{-1-\sqrt3}{2\sqrt2} +2\pi n \leq x \leq \frac{5\pi}{4} +arcsin \frac{-1-\sqrt3}{2\sqrt2}+2\pi n$

$b)\quad sinx-cosx \geq \frac{-1+\sqrt3}{2} \; |:\sqrt2\\\\sin(x-\frac{\pi}{4}) \geq \frac{-1+\sqrt3}{2\sqrt2}\; ,\; \; \frac{-1+\sqrt3}{2\sqrt2} \approx 0,26\\\\ \pi -arcsin \frac{-1+\sqrt3}{2\sqrt2} +2\pi n\leq x-\frac{\pi}{4} \leq arcsin \frac{-1+\sqrt3}{2\sqrt2} +2\pi n,\; n\in Z\\\\ \frac{5\pi}{4}-arcsin \frac{-1+\sqrt3}{2\sqrt2} +2\pi n\leq x \leq \frac{\pi}{4}+arcsin\frac{-1+\sqrt3}{2\sqrt2}+2\pi n$

$Otvet:\; x\in [\, \frac{\pi}{4}+arcsin \frac{-1-\sqrt3}{2\sqrt2} +2\pi n,\frac{5\pi}{4}+arcsin \frac{-1-\sqrt3}{2\sqrt2} +2\pi n]\cup$

$\cup [\, \frac{5\pi}{4} -arcsin \frac{-1+\sqrt3}{2\sqrt2}+2\pi n,\; \frac{\pi}{4} +arcsin \frac{-1+\sqrt3}{2\sqrt2} +2\pi n\, ]$

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