Question

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

Answer 1

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

$2)\; sin4x\cdot cos2x=sin2x\cdot cos4x\\\\sin4x\cdot cos2x-sin2x\cdot cos4x=0\\\\sin(4x-2x)=0\\\\2x=\pi n,\; \; x=\frac{\pi n}{2}\; ,\; n\in Z$

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