Sin2x+2cos^x-2=2cos2x, на промежутке от [2;5]. Created by acskirillacs. algebra-ru

Sin2x+2cos^x-2=2cos2x, на промежутке от [2;5]...

acskirillacs @acskirillacs

July 2022 1 5 Report

Sin2x+2cos^x-2=2cos2x, на промежутке от [2;5]

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NNNLLL54

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$sin2x+2cos^2x-2=2cos2x\; ,\; \; \; x\in [\, 2,5\, ]\\\\2sinx\cdot cosx-2(1-cos^2x)=2(cos^2x-sin^2x)\\\\2sinx\cdot cosx-2sin^2x-2cos^2x+2sin^2x=0\\\\2sinx\cdot cosx-2cos^2x=0\\\\2cosx\cdot (sinx-cosx)=0\\\\1)\quad cosx=0\; ,\; x=\frac{\pi}{2}+\pi n,\; n\in Z\\\\2)\quad sinx-cosx=0\, |:cosx\ne 0\\\\tgx-1=0\; ,\; \; \; tgx=1\\\\x=\frac{\pi}{4}+\pi k,\; k\in Z\\\\3)\quad x\in [\, 2,5\, ]\; ,\; 2rad\approx 114,5^\circ \; ;\; 5rad\approx 286,5^\circ \\\\x=\frac{\pi}{4}+\pi =\frac{5\pi}{4}=225^\circ$

$x=\frac{\pi}{2}+\pi =\frac{3\pi}{2}=270^\circ\\\\Otvet:\quad \frac{5\pi}{4}\; ;\; \frac{3\pi}{2}\; .$

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