Question

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

Answer 1

$sin2x+2cos^2x-2=2cos2x; ,; ; ; xin [, 2,5, ]\\2sinxcdot cosx-2(1-cos^2x)=2(cos^2x-sin^2x)\\2sinxcdot cosx-2sin^2x-2cos^2x+2sin^2x=0\\2sinxcdot cosx-2cos^2x=0\\2cosxcdot (sinx-cosx)=0\\1)quad cosx=0; ,; x=frac{pi}{2}+pi n,; nin Z\\2)quad sinx-cosx=0, |:cosxne 0\\tgx-1=0; ,; ; ; tgx=1\\x=frac{pi}{4}+pi k,; kin Z\\3)quad xin [, 2,5, ]; ,; 2radapprox 114,5^circ ; ;; 5radapprox 286,5^circ \\x=frac{pi}{4}+pi =frac{5pi}{4}=225^circ$

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