Question

A)2cos²-5sin+1=0
B) sin4scos2x=sin2xcos4x
C) cos2x-sinx=0

Answer 1

$1); 2cos^2x-5sinx+1=0\\2(1-sin^2x)-5sinx+1=0\\2sin^2x+5sinx-3=0\\(sinx)_1=-3 textless -1; ; net; reshenij,\\ (sinx)_2=frac{1}{2}; ,; ; x=(-1)^{m}frac{pi}{6}+pi m; ,; min Z; -; otvet.$

$2); sin4xcdot cos2x=sin2xcdot cos4x\\sin4xcdot cos2x-sin2xcdot cos4x=0\\sin(4x-2x)=0\\2x=pi n,; ; x=frac{pi n}{2}; ,; nin Z$

$3); cos2x-sinx=0\\(1-2sin^2x)-sinx=0\\2sin^2x+sinx-1=0\\(sinx)_1=-1; ,; x=(-1)^{n}cdot (-frac{pi}{2})+pi n=(-1)^{n+1}frac{pi}{2}+pi n,; nin Z\\(sinx)_2=frac{1}{2}; ,; x=(-1)^{k}cdot frac{pi}{6}+pi k; ,; kin Z$