Question

sin2x+sinx+2cosx=cos2x

Answer 1

$sin2x+sinx+2cosx=cos2x\\2sinxcdot cosx-(cos^2x-sin^2x)+sinx+2cosx=0\\2sinxcdot cosx-(1-sin^2x)+sin^2x+sinx+2cosx=0\\(2sinxcdot cosx+2cosx)+(2sin^2x+sinx-1)=0\\, [, 2sin^2x+sinx-1=0; ,; D=1+8=9,\\(sinx)_1=frac{-1-3}{4}=-1,; ; (sinx)_2=frac{-1+3}{4}=frac{1}{2}; ]\\2cosx(sinx+1)+2(sinx+1)(sinx-frac{1}{2})=0\\2(sinx+1)(cosx+sinx-frac{1}{2})=0\\a); ; sinx+1=0; ,; sinx=-1; ,; underline {x=-frac{pi}{2}+2pi n; ,; nin Z}\\b); ; cosx+sinx=frac{1}{2}; |:sqrt2$

$frac{1}{sqrt2} cosx+frac{1}{sqrt2}sinx=frac{1}{2sqrt2}\\sinfrac{pi}{4}cosx+cosfrac{pi}{4}sinx=frac{1}{2sqrt2}\\sin(x+frac{pi}{4})=frac{1}{2sqrt2}\\x+frac{pi}{4}=(-1)^{k} arcsinfrac{1}{2sqrt2}+pi k; ,\\underline {x=(-1)^{k}arcsinfrac{1}{2sqrt2}-frac{pi}{4}+pi k; ,; kin Z}$