Question

5-2sin(x)*cos(x)-5(sin(x)-cos(x)=0

Answer 1

$5-2, sinxcdot cosx-5(sinx-cosx)=0\\Zamena:; ; t=sinx-cosx; ,\\t^2=(sinx-cosx)^2=sin^2x+cos^2x-2, sinxcdot cosx=1-2. sinxcdot cosx; to \\2, sinxcdot cosx=1-t^2\\5-(1-t^2)-5t=0\\5-1+t^2-5t=0\\t^2-5t+4=0; ; ,; ; t_1=1; ,; t_2=4; ; (teorema; Vieta)\\a); ; sinx-cosx=1; ; to ; ; sinx-sin(frac{pi }{2}-x)=1; ,\\2sinfrac{x-(frac{pi}{2}-x)}{2}cdot cosfrac{x+(frac{pi}{2}-x)}{2}=1\\2sin(x-frac{pi}{4})cdot cosfrac{pi}{4}=1\\2sin(x-frac{pi }{4})cdot frac{sqrt2}{2}=1$

$sin(x-frac{pi}{4})=frac{sqrt2}{2}\\x-frac{pi}{4}=(-1)^{n}cdot frac{pi}{4}+pi n; ,; nin Z\\x=frac{pi }{4}+(-1)^{n}cdot frac{pi }{4}+pi n=left [ {{pi n; ,; esli; n=2k-1; ,; kin Z} atop {frac{pi}{2}+pi n; ,; esli; n=2k; ,; kin Z}} right. \\b); ; sinx-cosx=4\\2sin(x-frac{pi }{4})cdot frac{sqrt2}{2}=4\\sin(x-frac{pi}{4})=2sqrt2>1; ; to ; ; net; kornej; ,; t.k.; |sin(x-frac{pi}{4})|leq 1\\Otvet:; ; x=frac{pi}{4}+(-1)^{n}cdotfrac{pi }{4}+pi n; ,; nin Z$

$star ; ; cosx=sin(frac{pi}{2}-x); ; star \\star sinx-siny=2sinfrac{x-y}{2}cdot cosfrac{x+y}{2}; ; star$